Answer to Question #96694 in Mechanics | Relativity for Abru

Question #96694
A basketballer is trying to make a shot in an international game from the 3 point line (6.75 m
away from the hoop). The basketball hoop is 3.05 m above the floor, the basketballer’s hands
are 1.95 m above the floor when they release the ball at an angle of 39° above the horizontal.
What velocity would the ball have to have been thrown with to reach and fall down into the
hoop? And if the ball has a mass of 625 g what was the ball’s maximum potential energy during
the throw?
1
Expert's answer
2019-10-17T11:15:26-0400

The equation of trajectory of the ball when it moves as projectile


y=y0+tan(θ)xg2v02cos2(θ)x2y=y_0+\tan(\theta)\cdot x-\frac{g}{2v_0^2\cos^2(\theta)}\cdot x^2

In our case we have an equation


3.05=1.95+tan(39)6.759.82v02cos2(39)6.7523.05=1.95+\tan(39^{\circ})\cdot 6.75-\frac{9.8}{2v_0^2\cos^2(39^{\circ})}\cdot 6.75^2

So, the initial velocity


v0=9.2m/sv_0=9.2 \:\rm m/s

The maximum height


ymax=y0+v02sin2(θ)2gy_{max}=y_0+\frac{v_0^2\sin^2(\theta)}{2g}

=1.95+9.22sin2(39)2×9.8=3.66m=1.95+\frac{9.2^2\sin^2(39^{\circ})}{2\times 9.8}=3.66\:\rm m

So, the ball’s maximum potential energy during the throw


Emax=mgymaxE_{max}=mgy_{max}

=0.625kg×9.8N/kg×3.66m=22.4J=0.625\:\rm kg\times 9.8\: N/kg\times 3.66\: m=22.4\:\rm J


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment