Projectile motion question ( basketball being thrown into a hoop)

[mihir23]

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.

 

[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.

 

[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.