Question

You are standing on a slope of 20∘ to the horizontal and are going to throw a ball at vi = 13 m/s up the incline 

If you throw the ball at θ = 45 ∘ with respect to the horizontal, at what distance up the incline from your feet does it land? Assume the ball leaves your hand directly overhead, h = 2.1 m above the slope.

Answer 1

Vertical component of velocity =$13 sin 45=9.19$

Horizontal component of velocity =$13 cos 45=9.19$

After time t,

$S_y=2.1+9.19t- \frac{1}{2} \times9.8t^2=2.1+9.19t-4.9t^2$

$S_x=9.19t$

$tan 20 = \frac{S_y}{S_x}$

$S_x = \frac{S_y}{tan20} \implies 9.19t= \frac{S_y}{tan20} \implies t= \frac{S_y}{9.91tan20}$

$3.345t=2.1+9.19t-4.9t^2$

$x_{1,\:2}=\frac{-5.845\pm \sqrt{5.845^2-4\left(-4.9\right)\cdot \:2.1}}{2\left(-4.9\right)}=-0.28917, 1.48203$

$S_x=9.19 \times 1.5 =13.785 m$