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

Answer:

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

$h = u t + \dfrac{1}{2}gt^2$

$10= 0 + \dfrac{1}{2}\times 9.8 \times t^2$

$t = \sqrt{\dfrac{20}{9.8}}$

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

$d = \sqrt{10^2+21.45^2}$

d = 23.67 m

distance of boat from the release point = 23.67 m