Question

A boy is standing bridge and releases a ball from a height of 10 m, while a boat is approaching at a constant 15 on a m/s. How far away must the boat be from the release point of the ball for the ball to land in the boat?

Answer 1

Step-by-step explanation:

given,

height of the bridge = 10 m

constant speed of boat = 15 m/s

distance of the boat from the release point = ?

initial velocity of ball = 0

time taken by the ball to reach down

`h = u t + (1)/(2)gt^2`

`10= 0 + (1)/(2)* 9.8 * t^2`

`t = \sqrt{(20)/(9.8)}`

t = 1.43 s

distance travel by the boat in 1.43 second

s = v × t

s = 15 × 1.43

s = 21.45 m

boat distance from the release point

`d = √(10^2+21.45^2)`

d = 23.67 m

distance of boat from the release point = 23.67 m