# Projectile motion question ( basketball being thrown into a hoop)

1. Jun 24, 2012

### mihir23

1. The problem statement, all variables and given/known data
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.

2. Relevant 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)

3. 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.

#### Attached Files:

• ###### physics question diagram.jpg
File size:
15 KB
Views:
441
2. Jun 24, 2012

### Xisune

0.9m is not maximum height, so u^2 is not 17.64, but you don't need to find max height anyways.

Start with the displacement equations for both components, then put those equations together to find time.

3. Jun 24, 2012

### TSny

Look's like you assumed that the vertical component of velocity is zero at the instant the ball goes through the hoop. You can't impose that condition.