Ball is 2.04 sec in the air after it will undergoes free fall, Max height 20.14 m and after 0.99 sec the ball 15m above the ground.
The most basic principles of an object's motion are described by kinematics equations of motion. These equations control an object's motion in 1D, 2D, and 3D. They make it simple to compute expressions like an object's position, velocity, or acceleration at different points in time.
At max height, V=0.
So:
v=u+at
0=20–9.8t
9.8t=20
t=2.04 secs…
Ball is 2.04 sec in the air after it will undergoes free fall.
s = ut + ½ at^2
=20* 2.04 - (9.8 * 2.04² / 2)
= 40.8 - 20.39
= 20.41 m
Max height 20.14 m
ball 15m above the ground:
v^2 = u^2 + 2 as.
= 20^2 - 2 * 9.8 * 15
= 400- 294
=106
v = [tex]\sqrt{106}[/tex] =10.3 m/s
v=u+at
10.3 = 20 -9.8 * t
t = (20 -10.3) /9.8 = 0.99 sec
at 0.99 sec the ball 15m above the ground.
To learn more about acceleration refer,
https://brainly.com/question/27821888
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