Question

A rock is thrown horizontally from a bridge with a velocity of 17.0 m/s. It takes the rock 3.0 s to strike the water below. What is the magnitude of the velocity of the rock just before it hits the water

Answer 1

Answer:

V= 33.98 m/s

Explanation:

Given that

Horizontal speed ,u= 17 m/s

Time taken by rockets to strike the water ,t= 3 s

We know that acceleration due to gravity ,g= 9.81 m/s²

There is no any acceleration in the horizontal direction that is why the horizontal veloity will remain constant.

In the vertical direction

vy = uy+ g t

Initial velocity in vertical direction is 0 m/s.

vy= 0+ 9.81 x 3

vy = 29.43 m/s

The resultant velocity

$V=\sqrt{v_y^2+u^2}$

$V=\sqrt{29.43^2+17^2}\ m/s$

V= 33.98 m/s