Question

A projectile is fired from 22mm above ground. The projectile travels with an initial velocity of 55meter per second from an angle of elevation of 60degrees. Find the horizontal distance the ball traveled at the time it reaches the ground.

Answer 1

Solution

For this case we can create the following plot:

We can use the following formula to find the time to reach the ground:

$y=y_o+v_{oy}t-\frac{1}{2}gt^2$

And we can assume that yo =0 and g= 9.8m/s^2 then we have this:

$0=22+55\cdot\sin 60\cdot t-\frac{1}{2}\cdot\frac{9.8m}{s^2}t^2$

And solving we have this quadratic expression:

$-4.9t^2+47.632t+22=0$

And then solving for t using the quadratic formula we got:

$t=\frac{-47.632\pm\sqrt[]{(47.632)^2-4\cdot(-4.9)\cdot(22)}}{2\cdot(-4.9)}$

Solving we got:

t= 10.162 s

And then we can find the horixontal distance with this formula:

$R=v_x\cdot t=55\cdot\cos 60\cdot10.162s=279.455m$