Physics

Use the equation [ D = 1/2 g t2 ] .

D = the distance of free-fall = 50 m

g = acceleration of gravity = 9.8 m/sec2

t = time in free-fall

D = 1/2 g t2

2D / g = t2

t = sqrt( 2D / g ) = sqrt( 100 / 9.8 ) = 3.1944 seconds(rounded)

The horizontal velocity that the ball has when it's launched doesn't appear

in the calculation, because it makes absolutely no difference in the result.

Any effects of air resistance have been ignored in arriving at this solution.

Ignoring air resistance:

The distance from the base of the cliff is irrelevant as the only force acting on the ball is gravity acting straight down, giving it acceleration due to gravity.

s = ut + 1/2 at^2

For this problem: u = 0 m/s, t = 4 s, a = g (= acceleration due to gravity)

No value is given for g, but as an answer to the nearest tenth is required (1 dp), I'll use a value for g to 2 dp so that I can then round the answer to 1 dp: I'll use g ≈ 9.81 m/s^2

→ s ≈ 0 m + 1/2 x 9.81 m/s^2 x (4 s)^2 = 78.48 m

Rounded to the nearest tenth → 78.5 m.

horizontal distance = speed x time s = vt

45 = 15 t

t = 3 seconds