[azn-gilr]

Hi! Can anyone help me solving this problem.
A 2.00 tall basketball player is standing on the floor 10.0 m from the basket. If he shoots the ball at a 40 degree angle with the horizontal, at what initial speed mus he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is 3.05 m.

[terryg752]

Suppose velocity is V

Horizontal component of velocity is V cos 40

Horizontal distance = 10, Horizontal Acceleration = 0

Time = t = 10/(V Cos 40)

Vertical component of Initial velocity = u = V sin 40
Vertical acceration = -g
time = t found above

Vertical distance = 1.05

Use formula for VERTICAL motion:

s = ut + 1/2 at^2 (where a is acceleration)

You will get an equation in V, which solve.

If you get 2 values, use the higher value.

[azn-gilr]

Hi! Help me with this problem
A daredevil on a motorcycle leaves the end of a ramp with a speed of 35 m/s. If his speed is 33 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.

I used energy conservation energy but I can't find the answer. Any suggestion??

[terryg752]

Clearly, Horizontal component of speed (constant) = 33

Let angle he makes with horizontal when leaving ramp = A

Horizontal component of speed = 35 Cos A = 33

Find A

Then find

Verticle component of speed = u = 35 Sin A

Final verticale velocity v = 0

acceleration = a = -g

Height = s

Find s using

v^2 - u^2 = 2as