A ball is thrown from a roof top with an initial speed of 78.4 m/s at an angle o

Question

A ball is thrown from a roof top with an initial speed of 78.4 m/s at an angle of 30 degree. The roof top is 50 m above the ground. Draw a diagram of the trajectory of the ball showing its flight path from the moment is thrown until it hits the ground. Assume it is thrown to the right. What is the velocity (magnitude and direction) of the ball as it strikes the ground? How high is the ball when it is at its highest point? How far to the right of the wall is the ball, when it reaches the highest point? What is the velocity of the ball at this point? How far to the right of the building is the ball as it hits the ground?

Explanation / Answer

horizontal velocity of the ball will remain constant

by third equation of motion

v^2 = u^2 + 2as

v^2 = (78.4 * sin(30))^2 - 2 * 9.8 * 50

vertical velocity v = 23.59 m/s

net velocity = sqrt(horizontal velocity^2 + vertical velocity^2)

net velocity = sqrt((78.4 * cos(30))^2 + 23.59^2)

a) net velocity = 71.877 m/s

at heighest point final vertical velocity will be 0 m/s

0 = (78.4 * sin(30))^2 - 2 * 9.8 * s

s = 78.4 m

heighest point = 78.4 + 50

b) heighest point = 128.4 m

time taken by ball to reach heighest point t

by first equation of motion

v = u + at

0 = 78.4 * sin(30) - 9.8 * t

t = 4 sec

horizontal distance covered in 4 sec = 78.4 * cos(30) * 4

c) horizontal distance covered = 271.585 m

velocity of the ball at this point = 78.4 * cos(30)

d) velocity of the ball at this point = 67.89 m/s

time taken by ball to reach ground t1

by second equation of motion

s = ut + 0.5 * at^2

50 = 78.4 * sin(30) * t1 - 0.5 * 9.8 * t1^2

t1 = 6.4 sec

horizontal distance covered in 6.4 sec = 78.4 * cos(30) * 6.4

e) horizontal distance covered = 434.53 m

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