Question #241552

A student is standing at the top of a 10m tall building when they kick a soccer ball off the roof at a speed of 15m/s, at an angle of 30o above the horizontal. The ball then lands on the ground, some distance from the building.


1
Expert's answer
2021-09-24T09:25:24-0400

g=9.8ms2g =9.8\frac{m}{s^2}

x0=0;y0=10x_0=0; y_0=10

v=15ms\vec v= 15\frac{m}{s}

α=30°\alpha=30\degree

v=vx+vy\vec v= \vec v_x+\vec v_y

vx=vcosα13ms\vec v_x=\vec v\cos\alpha\approx13\frac{m}{s}

vy=vsinα=7.5ms\vec v_y=\vec v\sin\alpha=7.5\frac{m}{s}

y=y0+vytgt22y =y_0+v_yt-\frac{gt^2}{2}

When the ball falls to the ground:\text{When the ball falls to the ground:}

y=0y=0

y0+vytgt22=0y_0+v_yt-\frac{gt^2}{2}=0

10+7.5t4.9t2=010+7.5t -4.9t^2=0

t=2.39st =2.39s

x=x0+vxt=132.39=31.07mx = x_0+\vec v_xt= 13*2.39=31.07 m

Answer: distance from the building to where the ball lands is 31.07m\text{Answer: distance from the building to where the ball lands is }31.07m


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