Help with physics question about equations of motion

Need help with a Higher physics question, would be grateful if someone could please explain the solution?

A stone is thrown vertically upwards with a speed of 20m/s from the edge of a cliff. It lands with a speed of 50m/s at the bottom of the cliff. The effect of air resistance on the stone may be neglected.
a) What is the speed of the stone at its highest point from the top of the cliff?
b) What is the total time the stone is in the air?
c) High high above the top of the cliff did the ball reach?
d) How high is the cliff?

Thanks and much appreciated :)

OK, it's clear you do not understand trajectory physics. You need to reread your assignment and this time pay attention to some of the definitions. Like, what is the definition for "highest point from"? It's...

a) The point where Vy = 0, the vertical speed is zero. And as the stone was thrown vertically, Vx = Ux = 0 as well. ANS. See what I'm getting at? By knowing the physics, you can give the answer without working one equation. It was your inability to answer this question by sight that led me to conclude you really need to learn trajectory physics.

b) The time to max height is found from Vy = 0 = Uy - g tup; so that tup = Uy/g = 20/10 = 2 seconds [I'm using g = 10 m/s^2 for simplicity.] The time down is found from Vy = g tdn; so that tdn = Vy/g = 50/10 = 5 sec. So T = tup + tdn = 7 seconds. ANS

c) The distance (height) the stone rose above the cliff is found from S = Vavg tup = Uy/2 tup = 20/2 * 2 = 20 meters. ANS. Note Vavg = (Vy + Uy)/2 = Uy/2 as Vy = 0.

d) The distance the stone fell is found from S = Vy/2 tdn = 25*5 = 125 meters of which H = 125 - 20 = 105 meters was the height of the cliff. ANS. Note again, Vavg = (Vy + Uy)/2 only this time Vy = 50 mps and Uy = 0 as the rock is falling here.

You must be familiar with the three basic equations of motion under constant acceleration: