A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10m s -2 , with what velocity will it strike the ground? After what time will it strike the Ground?

Height,h= 20m

"gently dropped" means initial velocity, u, =0m/s

Uniform acceleration, g=10m/s²

using the formula v²=u²+2gs

Substituting, v²=0²+2*10*20

v²=400

v=

v=20m/s

Time, t at which is strike the ground

using the formula v=u+at

Substituting, 20=0+10t

20=10t

t=2s

This means that the bank strike the ground with a velocity of 20m/s in 2s.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14