Question

Determine the speed at which the basket ball at A must be thrown at an angle 30o so that if makes it to the basket at B.
Also find at what speed it passes through the hoop.

Answer 1

Given : θ=30o
To find : Speed at which basket ball must be thrown
Solution :
Assume that the basket ball be thrown with initial velocity u and it takes time t to reach B
HORIZONTAL MOTION
Here the velocity is constant
8 = ucos30 x t
`t=8/(ucos30)=9.2376/u`……..(1)
vB = ucos30 (Since velocity is constant in horizontal motion) ………(2)
VERTICAL MOTION
Initial vertical velocity (uv) = usin30 =0.5u …….(3)
Vertical displacement(s) = 2.4 - 1.2 = 1.2
`t=9.2376/u`
Using kinematical equation :
s = ut + `1/2` x at²
`1.2=u/2x9.2376/u-1/2x9.81x((9.2376)/u)^2`
u²=122.4289
u=11.0648 m/s
u_v=0.5u (From 3)
u_v = 0.5 x 11.0648
= 5.5324 m/s
Using kinematical equation
v_v² = u_v² + 2as
v_v² = 5.53242 - 2 x 9.81 x 1.2
v_v = 2.6622 m/s
v_h = 11.0648cos30 = 9.5824 m/s (From 2)
`v_B=sqrtV_v^2+v_h^2`
v_B = 9.9441 m/s

`apha=tan^-1((2.6577)/(9.5824))`
= 15.5015°
Speed at which the basket-ball at A must be thrown = 11.0648 m/s (30° in first quadrant)
Speed at which the basket-ball passes through the hoop = 9.9441 m/s(15.5015° in fourth quadrant)