Let u = initial vertical velocity.
Assume that
g = 9.81 m/s²,
Wind resistance is ignored.
When t = 0.220 s, the height is h = 0.537 m. Therefore
0.537 m = (u m/s)*(0.220 s) - (1/2)*(9.81 m/s²)*(0.220 s)²
0.537 = 0.22u - 0.2372
u = 3.519 m/s
The upward velocity after 0.220 s is
v = 3.519 - 9.81*0.22 = 1.363 m/s
At maximum height, the upward velocity is zero. The maximum height, H, is given by
(3.519 m/s)² - 2*(9.81 m/s²)*(H m) = 0
12.3834 - 19.6H = 0
H = 0.632 m
It goes higher by 0.632 - 0.537 = 0.095 m
Answers:
(a) The initial speed is 3.519 m/s.
(b) The speed at 0.537 m height is 1.363 m/s.
(c) It goes higher by 0.095 m.
Your question has no context, nobody’s going to be able to answer it.
Using the kinematic equation below we can determine the distance traveled if t=2, a=7.4m/s^2. First we must determine the final velocity:

Now we will determine the distance traveled:

Therefore, the drag racer traveled 81.83 meters in 2 seconds.
Answer:
Acceleration = 10.06 m/s²
Explanation:
1 mile = 1.6093km
1609.3m = 1 mile
1 m =
mile
50.0 miles/hour =
m/s
= 22.35m/s
from equation
S = Ut + 1/2 at²
v = U + at
22.35 = 0 + a * 2.22
a = 22.35 ÷ 2.22
= 10.06 m/s²