Question

a football is thrown upward at a 31° angle to the horizontal.the acceleration of gravity is 9.8m/s^2. To throw the ball a distance 77 m, what must be the initial speed of the ball. Answer in units of m/s

Answer 1

The ball's vertical position $y$ in the air at time $t$ is

$y=v_0\sin31^\circ\,t-\dfrac g2t^2$

The ball is at its original height when $y=0$, which happens at

$v_0\sin31^\circ\,t-\dfrac g2t^2=\dfrac t2\left(2v_0\sin31^\circ-gt)=0$

$\implies t=0\text{ and }t=\dfrac{2v_0\sin31^\circ}g$

Meanwhile, the ball's horizontal position $x$ at time $t$ is

$x=v_0\cos31^circ\,t$

So when the ball reaches its original height a second time, the ball will have traveled a horizontal distance of

$x=\dfrac{2{v_0}^2\sin31^\circ\cos31^\circ}g=\dfrac{{v_0}^2\sin(2\cdot31^\circ)}g$

(which you might recognize as the formula for the range of a projectile)

To reach a distance of $x=77\,\rm m$, the initial speed $v_0$ would be

$77\,\mathrm m=\dfrac{{v_0}^2\sin62^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\implies v_0=29\dfrac{\rm m}{\rm s}$