Easy physics question, just don't understand the question!

Question

full workings with explanation please, and thanks in advance! :)

---------- A ball is projected vertically upwards from the top of a building which is 300m high. The speed of the ball after 1s is 10m/s. Take g = 10 m/s^2 and neglecting air resistance, calculate: a) the time it takes to reach maximum height b) the maximum height measured from the ground reached by the ball c) the speed of the ball when it reaches the ground

Answer 1

Start by finding the initial vertical speed from

 

v(t) = v₀ + at = v₀ - gt

 

g = 10 m/s²

 

v(1) = 10 m/s

 

10 = v₀ - 10(1) ⇒

 

v₀ = 20 m/s²

 

(a)

 

At maximum height, the instantaneous vertical speed is zero.

 

v₀ - gt = 0 ⇒

 

t = v₀/g = 20/10 s = 2 s

 

(b)

 

h(t) = -gt²/2 + v₀t + h₀

 

h(t) = -5t² + 20t + 300,

 

with h in meters, t in seconds

 

Plug into h(t) the time calculated in part (a) to get h_max.

 

(c)

 

From conservation of energy,

 

mv₀²/2 + mgh₀ = mv²/2 + mg(0)

 

v₀²/2 + gh₀ = v²/2

 

v = √(v₀² + 2gh₀)

 

Plug in

 

v₀ = 20

g = 10

h₀ = 300

 

and get v in m/s.  You can finish from here.