Question

A ball is thrown horizontally from a height of 19.1 m and hits the ground with a speed that is 5.00 times its initial speed. What is the speed in the vertical direction just before the ball hits the ground?

Answer 1

let, initial velocity = v

using energy conservation,

KEi + PEi = KEf + PEf

(1/2)*m*v^2 + m*g*h = (1/2)*m*(5v)^2 + 0

v = g*h / sqrt(12)

v = 9.8*19.1 / sqrt(12)

v = 3.95 m/s

let us assume that the final velocity has an angle $\Theta$ with horizontal.

horizontal component of final velocity will be equal to initial velocity because no external force is there

5v*cos $\Theta$ = v

$\Theta$ = 78.46 deg

so speed in the vertical direction is

vy = 5v*sin $\Theta$

vy = 5*3.95*sin(78.46)

vy = 19.35 m/s

answer