Solutions and Explanations to Projectile Problems

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Solutions and detailed explanations to projectile problems are presented . These solutions may be better understood when projectile equations are first reviewed.


Problem 1: An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.

a) What is the maximum height reached by the object?

b) What is the total flight time (between launch and touching the ground) of the object?

c) What is the horizontal range (maximum x above ground) of the object?

d) What is the magnitude of the velocity of the object just before it hits the ground?

Solution to Problem 1:

projectiles
a) The formulas for the components Vx and Vy of the velocity and components x and y of the displacement are given by

Vx = V0 cos(θ)       Vy = V0 sin(θ) - g t

x = V0 cos(θ) t       y = V0 sin(θ) t - (1/2) g t2

In the problem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2.

The height of the projectile is given by the component y, and it reaches its maximum value when the component Vy is equal to zero. That is when the projectile changes from moving upward to moving downward.(see figure above) and also the animation of the projectile.

Vy = V0 sin(θ) - g t = 0

solve for t

t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds

Find the maximum height by substituting t by 0.86 seconds in the formula for y

maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 meters

b) The time of flight is the interval of time between when projectile is launched: t1 and when the projectile touches the ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence

V0 sin(θ) t - (1/2) g t2 = 0

Solve for t

t(V0 sin(θ) - (1/2) g t) = 0

two solutions

t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g

Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.

c) In part c) above we found the time of flight t2 = 2 V0 sin(θ) / g. The horizontal range is the horizontal distance given by x at t = t2.

range = x(t2) = V0 cos(θ) t2 = 2 V0 cos(θ) V0 sin(θ) / g = V02 sin(2θ) / g = 202 sin (2(25°)) / 9.8 = 31.26 meters


d) The object hits the ground at t = t2 = 2 V0 sin(θ) / g (found in part b above)

The components of the velocity at t are given by

Vx = V0 cos(θ)       Vy = V0 sin(θ) - g t

The components of the velocity at t = 2 V0 sin(θ) / g are given by

Vx = V0 cos(θ) = 20 cos(25°)       Vy = V0 sin(25°) - g (2 V0 sin(25°) / g) = - V0 sin(25°)

The magitude V of the velocity is given by

V = √[ Vx2 + Vy2 ] = √[ (20 cos(25°))2 + (- V0 sin(25°))2 ] = V0 = 20 m/s


Problem 2: A projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. The projectile hits the incline plane at point M.

a) Find the time it takes for the projectile to hit the incline plane.

b)Find the distance OM.

projectile over incline plane

Solution to Problem 2:

a) The x and y components of the displacement are given by

x = V0 cos(θ) t       y = V0 sin(θ) t - (1/2) g t2

with θ = 22 + 10 = 32° and V0 = 15 m/s

The relationship between the coordinate x and y on the incline is given by

tan(10°) = y / x

Substitute x and y by their expressions above to obtain

tan(10°) = ( V0 sin(θ) t - (1/2) g t2) / V0 cos(θ) t

Simplify to obtain the equation in t

(1/2) g t + V0 cos(θ) tan(10°) - V0 sin(θ) = 0

Solve for t

t =
 
V0 sin(θ) - V0 cos(θ) tan(10°)
0.5 g
=
 
15 sin(32°) - 15 cos(32°) tan(10°)
0.5 (9.8)
= 1.16 s


b)

OM = √[ (V0 cos(θ) t)2 + ( V0 sin(θ) t - (1/2) g t2)2 ]

OM (t=1.16)= √[ (15 cos(32) 1.16)2 + ( 15 sin(32) 1.16 - (1/2) 9.8 (1.16)2)2 ] = 15 meters


Problem 3: A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.

a) What is the range of values of the initial velocity so that the projectile falls between points M and N?

projectiles over pond

Solution to Problem 3:

a) The range is given by x = V02 sin(2θ) / g

We want to have the range greater than OM and smaller that ON, with OM = 10 + 20 = 30 m and ON = 10 + 20 + 10 = 40 m

30 < V02 sin(2θ) / g < 40

30 g / sin(2θ) < V02 < 40 g / sin(2θ)

√ [ 30 g / sin(2θ) ] < V0 < √ [ 40 g / sin(2θ) ]

18.4 m/s < V02 < 21.2 m/s


Problem 4:A ball is kicked at an angle of 35° with the ground.

a) What should be the initial velocity of the ball so that it hits a target that is 30 meters away at a height of 1.8 meters?

b) What is the time for the ball to reach the target?

projectiles ball

Solution to Problem 4:

a)

x = V0 cos(35°) t

30 = V0 cos(35°) t

t = 30 / V0 cos(35°)

1.8 = -(1/2) 9.8 (30 / V0 cos(35°))2 + V0 sin(35°)(30 / V0 cos(35°))

V0 cos(35°) = 30 √ [ 9.8 / 2(30 tan(35°)-1.8) ]

V0 = 18.3 m/s

b)


t = x / V0 cos(35°) = 2.0 s



Problem 5: A ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters.

a) What is the size of angle θ?

b) What is time of flight of the ball?

Solution to Problem 5:

a)

Let T be the time of flight. Two ways to find the time of flight

1) T = 200 / V0 cos(θ) (range divided by the horizontal component of the velocity)

2) T = 2 V0 sin(θ) / g (formula found in projectile equations)

equate the two expressions

200 / V0 cos(θ) = 2 V0 sin(θ) / g

which gives

2 V02 cos(θ)sin (θ) = 200 g

V02 sin(2θ) = 200 g

sin(2θ) = 200 g / V02 = 200 (9.8) / 602

Solve for θ to obtain

θ = 16.5°

b)

Time of flight = 200 / V0 cos(16.5°) = 3.48 s


Problem 6: A ball of 600 grams is kicked at an angle of 35° with the ground with an initial velocity V0.

a) What is the initial velocity V0 of the ball if its kinetic energy is 22 Joules when its height is maximum?

b) What is the maximum height reached by the ball

Solution to Problem 6:

a)

When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ).

22 = (1/2) m (Vx)2

22 = (1/2) 0.6 (V0 cos (35°))2

V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s

b)

Initial kinetic energy (just after the ball is kicked)

Ei = (1/2) m V02 = (1/2) 0.6 (10.4)2 = 32.4 J

The difference between initial kinetic energy and kinetic energy when the ball is at maximum height H is equal to gain in potential energy

32.4 - 22 = m g H

H = 10.4 / (0.6 * 9.8) = 1.8 m


Problem 7: A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.

a) What is the size of the angle θ?

b) At what initial velocity was the projectile launched?

Solution to Problem 7:

a)

Vx = V0 cos (θ) = 1000 / 40 = 25 m/s

Time of flight = 2 V0 sin (θ) / g (formula found in projectile equations)

= 40 s

V0 sin (θ) = 20 g

Combine the above equation with the equation V0 cos (θ) = 25 m/s found above to write

tan (θ) = 20 g / 25

Use calculator to find θ = 82.7 °

b)

We now use any of the two equations above to find V0.

V0 cos (θ) = 25 m/s

V0 = 25 /cos (82.7 °) = 196.8 m/s


Problem 8: The trajectory of a projectile launched from ground is given by the equation y = -0.025 x2 + 0.5 x, where x and y are the coordinate of the projectile on a rectangular system of axes.

a) Find the initial velocity and the angle at which the projectile is launched.

Solution to Problem 8:

a)

y = tan(θ) x - (1/2) (g / (V0 cos (θ))2 ) x2 (formula found in projectile equations)

hence tan(θ) = 0.5 which gives θ = arctan(0.5) = 26.5 °

-0.025 = -0.5 (9.8 / (V0 cos (26.5 °))2 )

Solve for V0 to obtain V0 = 15.6 m/s


Problem 9: Two balls A and B of masses 100 grams and 300 grams respectively are pushed horizontally from a table of height 3 meters. Ball has is pushed so that its initial velocity is 10 m/s and ball B is pushed so that its initial velocity is 15 m/s.

a) Find the time it takes each ball to hit the ground.

b) What is the difference in the distance between the points of impact of the two balls on the ground?

Solution to Problem 9:

a) The two balls are subject to the same gravitational acceleration and therefor will hit the ground at the same time t found by solving the equation

-3 = -(1/2) g t2

t = √ (3(2)/9.8) = 0.78 s

b) Horizontal distance XA of ball A

XA = 10 m/s * 0.78 s = 7.8 m

Horizontal distance XB of ball B

XB = 15 m/s * 0.78 s = 11.7 m

Difference in distance XA and XB is given by

|XB - XA| = |11.7 - 7.8| = 3.9 m




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Updated: 26 September 2015

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