how do you find initial speed?

Question

This is the question I have:
A ball is thrown at an angle of 45* to the ground. If the ball lands 90m away, what was the initial speed of the ball?

If someone could explain how to do this I would apreciate it. You don't have to give me the answer, just how to solve it. Thanks

Answer 1

You use the equation that
x(t)=v0*cos(45)*t
where t is the flight time and v0 is the initial speed
you are given that x(t)=90 when the ball lands.

Ignoring air friction, the ball will land in twice the time it takes to reach apogee.

Apogee time is found by taking v(t)=0
where
v(t)=v0*sin(45)-g*t
or t=v0*sin(45)/g at apogee
and 2*sin(45)v0/g when the ball lands.

Plugging that t into x(t), we find that
90=2*v0^2*cos(45)*sin(45)/g

solve for v0
29.7 m/s using g=9.81


j

Answer 2

If the ball is projected with a velocity 'u' at angle 45, then component of its velocity in horizontal direction is 'u cos45' and component of its velocity in vertical direction is 'u sin45' .

The horizontal motion is uniform but vertical motion is with acceleration 'g' due to gravity in vertical downward direction.

The ball covers 90 m with uniform velocity 'u cos45' in time 't' in horizontal direction , therefore,
t=90 / u cos45--------(1)

The vertical displacement of the ball in time 't' is zero,

using the eqn " s=ut+(1/2)at^2 "

zero= (u sin45) t-(1/2)gt^2------------(2)

substituting 't' from eqn (1) in eqn (2).
zero= 90 -(1/2)g90x90/(ucos45)^2

solving this eqn,
we get 'u'=sq rt g x90=sq rt 9.8x90=29.69 m/s
The ball is projected with a speed of 29.69 m/s