Answer to Question #173664 in Mechanics | Relativity for Daniella

Question #173664

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.


1
Expert's answer
2021-03-23T08:08:23-0400

Vertical component of velocity =13sin45=9.1913 sin 45=9.19

Horizontal component of velocity =13cos45=9.1913 cos 45=9.19

After time t,

Sy=2.1+9.19t12×9.8t2=2.1+9.19t4.9t2S_y=2.1+9.19t- \frac{1}{2} \times9.8t^2=2.1+9.19t-4.9t^2


Sx=9.19tS_x=9.19t


tan20=SySxtan 20 = \frac{S_y}{S_x}


Sx=Sytan20    9.19t=Sytan20    t=Sy9.91tan20S_x = \frac{S_y}{tan20} \implies 9.19t= \frac{S_y}{tan20} \implies t= \frac{S_y}{9.91tan20}


3.345t=2.1+9.19t4.9t23.345t=2.1+9.19t-4.9t^2


x1,2=5.845±5.84524(4.9)2.12(4.9)=0.28917,1.48203x_{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


Sx=9.19×1.5=13.785mS_x=9.19 \times 1.5 =13.785 m


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