Apiece of stone is thrown up from a height of 25 meters above the ground. after 2 seconds the stone reaches the highest point.
a = calculate the starting speed
b = calculate the maximum height from the ground
c = calculate at what time the stone hits the ground
d = calculate the speed of the stone on time = 4 seconds
can someone me solve this

A. 19.6 m/s

B. 44.6 m

C. 5.0 s

D. -19.6 m/s

Explanation:

At the highest point, the stone's velocity is 0.

A. Given:

v = 0 m/s

t = 2 s

a = -9.8 m/s²

Find: v₀

v = at + v₀

0 = (-9.8)(2) + v₀

v₀ = 19.6 m/s

B. Given:

y₀ = 25 m

t = 2 s

v₀ = 19.6 m/s

a = -9.8 m/s²

Find: y

y = y₀ + v₀ t + ½ at²

y = 25 + (19.6)(2) + ½(-9.8)(2)²

y = 44.6 m

C. Given:

y₀ = 25 m

y = 0 m

v₀ = 19.6 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 25 + (19.6)t + ½(-9.8)t²

0 = 4.9t² − 19.6t − 25

t ≈ 5.0 s

D. Given:

v₀ = 19.6 m/s

a = -9.8 m/s²

t = 4 s

Find: v

v = at + v₀

v = (-9.8)(4) + 19.6

v = -19.6 m/s

The velocity is -19.6 m/s.  If you want the speed, or magnitude of the velocity, take the absolute value: 19.6 m/s.

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