Question

A stone is thrown from the ground upward with an initial velocity of 20 m/s

a- At what time the stone will reach the maximum height?
b- What is the maximum height the stone can reach ?
c- What is the height of the stone from the ground after 3 seconds ?

I solved ( A) and ( B) and I need answer for part ( C )

Answer
A- Vf = Vi + gt
0 = 20 + ( 9.8 m/s2) t
20 + 9.8t = 0
9.8 t = -20 so t = -20/9.8 m/s^2 = 2.04 second

B- Vf^2 = Vi^2 + 2g ( Xf – Xi)
(0)^2 = (20)^2 + 2 ( -9.8 m/s^2) ( Xf – 0)
(20)^2 + 2 (-9.8 m/s^2)(Xf) = 0
400 - 19.6 (Xf) = 0
-19.6 Xf = -400
So Xf = - 400/ -19.6 = 20.4 m

Answer 1

c) use the formula: h = h₀ + v₀t - gt²/2.
for h₀ = 0, v₀ = 20 m/s, g = 10 m/s², t = 3s:
h(t = 3) = 20*3 - 10*9/2 = 15 m.