Astone is thrown up from a platform with an initial velocity of 19.8 m/s. if
the top of the platform is taken to be the ground level, calculate, (a) the time
taken to reach the maximum height, (b) the maximum height reach by the
stone (c) its velocity just before reaching the ground and (d) total time taken
to reach back to the top of the platform

a) 2.02 s

b) 20.0 m

c) -19.8 m/s

d) 4.04 s

Explanation:

Given:

v₀ = 19.8 m/s

a = -9.8 m/s²

a) Find t when v = 0 m/s.

v = at + v₀

(0 m/s) = (-9.8 m/s²) t + (19.8 m/s)

t = 2.02 s

b) Find Δy when v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (19.8 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 20.0 m

c) Find v when Δy = 0 m.

v² = v₀² + 2aΔy

v² = (19.8 m/s)² + 2 (-9.8 m/s²) (0 m)

v = -19.8 m/s

d) Find t when Δy = 0 m.

Δy = v₀ t + ½ at²

(0 m) = (19.8 m/s) t + ½ (-9.8 m/s²) t²

t = 4.04 s

i think its c if i'm wrong