Projectile A is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface below, 3.0 seconds later. Projectile B is launched horizontally from the same location at a speed of 30 meters per second. How long does it take Projectile B to reach the ground?

Answer: 3 seconds

Explanation:

Initial velocity (u) of projectile A in vertical direction = 0m/s

acceleration due to gravity a=g=9.81m/s^2

Time taken (t) of projectile A = 3s

Initial velocity of projectile B = 0m/s (vertical direction)

We can get height of cliff using parameters of projectile A since it's the same location.

Height (S) = u*t + 0.5*a*t^2

u = 0

S = 0.5*9.81*3^2 = 44.145m

Time taken for projectile B to reach the ground:

S = u*t + 0.5*a*t^2

u = 0, S=44.145m, a=9.81m/s^2

44.145 = 0.5*9.81*t^2

44.145 = 4.905*t^2

44.145 : 4.905 = t^2

9 = t^2

t = sqrt (9)

t = 3seconds