Question

A basketball player shooting the ball from a height of 2.50 m with an elevation of 37 ° and scores in the basket located at 6.25 m distance and 3.05 m in height. Calculate the rate at which released the ball.

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Answer 1

@sami-21

Answer 2

y=H+v0*sin*t-1/2*g*t2

vy=v0*sin-g*t
ay=-g

yeh formulas use ho rhy hain
lekin muje samaj nhi aa rhi solve kese karon,,

Answer 3

@Muskan
here is diagram which may explain a bit to you the apparent height would be 0.55m distance(range)=6.25 angle =37' i think you can caculate initial velocity