Abaseball is launched horizontally from a height of 1.8 m. the baseball travels 0.5 m before hitting the ground. how fast is the baseball moving, rounded to the nearest hundredth? m/s

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.

this would be about 43%

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y