Projectile Motion/Physics Problem Help!!!?

Question

A basketball player who is 2.00m tall is standing on the floor 10.0m from the basket. if he shoots the ball at a 40 degree angle with the horizontal, at what initial speed must he throw the ball so that it goes through the hoop without striking the backboard? the basket height is 3.05m.

Answer 1

Unfortunately, this needs a calculator. However, I'll give some direction as to how to go about it.

Firstly, it is always good to make a sketch of the problem.

Secondly, you must be familiar with the projectile motion equations. For this problem we will make use of these,

X = UT, V = UT + GT, Y = VT + 1/2GT^2

Substitute the available information, and not forgetting to include cos and sine in the equation - you will get the desired solution.

N.B: Y = 3.05 - 2 = 1.05 m

Answer 2

Okay. so the ball falls -10m/s2 below what the theoretical height would be without gravity. Tan40=.839 which gives the slope. Multiply by 10, and it would rise 8.39m upward within 10m without G. That means it must move such that there is a 8.39-3.05= 5.34m fall below that point. so now it's known that it should be about 10m/s. D=1/2at2= 5.34=1/2(10)t2= t=1.03s. It has to move the horizontal distance of 10m in 1.03s = 9.709m/s.
phew.