# Answer on Mechanics | Relativity Question for zizi

Question #3131

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

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

Expert's answer

c) use the formula: h = h

for h

h(t = 3) = 20*3 - 10*9/2 = 15 m.

_{0}+ v_{0}t - gt^{2}/2.for h

_{0}= 0, v_{0}= 20 m/s, g = 10 m/s^{2}, t = 3s:h(t = 3) = 20*3 - 10*9/2 = 15 m.

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