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?
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 with horizontal.
horizontal component of final velocity will be equal to initial velocity because no external force is there
5v*cos = v
= 78.46 deg
so speed in the vertical direction is
vy = 5v*sin
vy = 5*3.95*sin(78.46)
vy = 19.35 m/s
answer
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