Question

A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assume no air resistance. How long until the ball reaches the ground?​

Answer 1

The horizontal motion has no effect on the vertical drop.

From a drop, the distance the ball falls in 'T' seconds is

D = 4.9 T^2

so

2.2 = 4.9 T^2

T^2 = 2.2/4.9

T^2 = 0.449 sec^2

T = 0.67 second

Answer 2

We have that A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assuming no air resistance will take t to reach the ground

t=0.670sec

From the question we are told that

A ball is thrown horizontally at a height of 2.2 meters

a velocity of 65m/s off a cliff

Generally the equation for the Time is mathematically given as

s=ut+1/2gt^2

Therefore

$2.2=0t+1/2(9.8)t^2\\\\t=\sqrt{\frac{2.2}{0.5*9.8}}\\\\t=0.670sec$

Therefore

A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assuming no air resistance will take t to reach the ground

t=0.670sec

For more information on this visit

https://brainly.com/question/12319416?referrer=searchResults