Question

A ball is thrown horizontally from a height of 10 m with velocity of 21 ms': How
far off it hit the ground and with what velocity?
(Ans: 30m, 25 ms​

Answer 1

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, the ball will fall 10m with an acceleration of g = 9.81m/s^2, and we want to find the time, t, required for this, so we use equation (1)

s = ut + ½at^2

10 = 0 + 4.905t^2

t = √(10/4.905) = 1.428s

and during this time, the ball is moving with a horizontal velocity of 21m/s, so it travels a distance of 1.428(21) = 29.99m

So, the horizontal velocity is 21m/s, and the vertical velocity is found from equation (3)

v = u + at, so v = 9.81(1.428) = 14.0m/s

Therefore, the ball strikes the ground with a velocity of √(21^2 + 14^2) which is 25.24m/s at an angle of Tan^-1(21/14) = 56.31°

So, the ball hits the ground 29.99m from the launch point, striking the ground with a velocity of 25.24m/s at an angle of 56.31°