Question

Ball 1 is thrown vertically upwards at 8 m/s from a height of 30 meters from the side a building. A second ball is thrown upwards from the ground (height of 0 m) 0.39 seconds after the ball 1 is thrown upwards. With what speed, in m/s, does the second ball need to be thrown such that it collides with ball 1 when ball 1 is 8 meters above the ground?

Answer 1

here,

for ball 1,

initial speed , u1 = 8 m/s

height , h1 = 30 m

final height , h2 = 8 m

let the time taken to reach h2 be t1

using seccond equation of motion

(h2 - h1) = u * t1 - 0.5 * g * t1^2

( 8 - 30) = 8 * t1 - 0.5* 9.81 * t1^2

solving for t1

t1 = 3.08 s

for ball 2

time taken , t2 = t1 + 0.39 s = 3.47 s

height , h2 = 8 m

let the initial speed of the ball2 be u2

h2 = u2 * t2 - 0.5 * g * t2^2

8 = u2 * 3.47 - 0.5 * 9.81 * 3.47^2

solving for u2

u2 = 19.3 m/s

the initial speed of ball 2 must be 19.3 m/s upwards