- #1

mihir23

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## Homework Statement

A basketball shoots a ball at an angle of 55° which is 4.3m away( horizontal dist). If the ball is released from a height of 2.1m and lands in the hoop, which is 3m high. Calculate the initial speed of the ball for this shot to be successful.

## Homework Equations

initial horiz velocity = v cos(θ)

initial vert velocity = v sin(θ)

Resultant initial velocity = Sqroot(initial horiz v ^2 + initial vert v^2)

## The Attempt at a Solution

I decided that the vertical displacement the ball travels is 0.9 m up(3-2.1m)

The question does not give the max height , so i was confused about how to approach it

What i did:

a = -9.8

u= ?

v^2 =u^2 +2as

u^2 =17.64

initial vert velocity = v sin(θ)

therefore

v(initial vert) = 4.2/sin 55

4.2 = v sin 55

= 5.127

v(initial horiz)=5.127/tan55

=3.59

total initial v = Sqroot(5.127^2 + 3.6^2)

=6.26 m/s

This answer is wrong according to the back of my book. Could you please tell me how i have done it incorrectly. If you guys really want it, then i'll post the book's answer.