A basketball player attempts to make a basket from distance

A basketball player attempts to make a basket from distance of 8m. the basket is 3 m high and the ball leaves his hands from height of 1m if he throws the ball at an angle of 45 degrees what does initial velocity of the ball need to be

Solution

Let initial velocity is v.

So horizontal velocity= v cos45, vertical velocity = v sin45

vertical distance travelled = 2 m

Horizontal distance travelled = 8m

Vertical accelration = -9.8

=horizontal accelration=0

Let time be t

fo for horizontal motion

s=ut+1/2at^2

8= vsin45*t;

so t= 8/vsin45

for vertical motion

s=ut+1/2at^2

s=2 t=8/vsin45 a= -9.8

2= vcos45*8/vsin45 +1/2 *(-9.8) * (8/vsin45)^2

v^2 = 9.8*8*8/(6*sin45*sin45*2)

v=10.22 m/s