Afrog jumps vertically upward from a 20m tall building with an initial velocity of 10m/s. how high above the ground will the frog reach? what is its velocity as it hits the ground below?

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.
Therefore,
u = 10 m/s, initial upward velocity.
H = - 20 m, position of the ground.
g = 9.8 m/s², acceleration due to gravity.

Part (a)
When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore
u² - 2gh = 0
h = u²/(2g) = 10²/(2*9.8) = 5.102 m
At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

25.1 m above ground

Part (b)
Let v = the velocity when the frog hits the ground. Then
v² = u² - 2gH
v² = 10² - 2*9.8*(-20) = 492
v = 22.18 m/s

The frog hits the ground with a velocity of 22.2 m/s

The answer is frictional force