Here we’re solving a problem where a ball is projected horizontally from a height of h=1.8 m with a horizontal velocity of Vx. At the impact with the ground, the ball has travelled 0.5 m horizontally.
Solution:
We will need kinematic equations
V1^2-V0^2=2aS ………………….(1)
S=V0*t + (1/2)at^2………………..(2)
Where
S=displacement (distance), m
V0=initial velocity, m/s
V1=final velocity, m/s
a=acceleration, m/s^2
t=time, seconds
At the point of impact, there a vertical velocity (downwards) of Vy.
The horizontal velocity Vx remains constant since projection till impact.
Vertical velocity Vy:
Using equation (1),
V0=0 (projected horizontally, so vertical velocity=0)
S=1.8 m (downwards)
a=9.81 m/s^2 (acceleration due to gravity, downwards)
=>
Vy=V1=sqrt(V0^2+2*a*S)=sqrt90+2*9.81*1.8)=5.9427
Horizontal velocity, Vx:
ball travelled 0.5m in time t it took ball to hit ground.
Using equation (2),
S=1.8m
V0=0
a=9.81
=>
1.8=0*t+(1/2)(9.81)t^2
Solve for t
t=sqrt(2*1.8/9.81)=0.60578 s
Horizontal velocity, Vx = 0.5/0.60578 = 0.82538 s
Speed of ball on impact is the vectorial sum of Vx and Vy:
Speed = sqrt(Vx^2+Vy^2)=sqrt(5.9427^2+0.82538^2)=5.99977 m/s, say 6.0 m/s.
