Question

A ball is thrown upward with initial velocity v0 = 15.0 m/s at an angle of 30° with the horizontal. The thrower stands near the top of a Jong hill which slopes downward at an angle of 20°. Determine how far down the slope the ball strikes.

Answer 1

Vo vector is 30^o above horizontal , and inclined hill(slope) is 20^o below horizontal

We need to find L as shown in figure,

We can solve this question by considering equations along the (incline) slope and perpendicular to the (incline)slope,

Taking horizontal along the slope, and verticle as perpendicular to slope,
Horizontal(along the slope) velocity is VoCos50 and Verticle velocity is VoSin50,
Acceleration due to gravity along the slope is gCos70 and perpendicular to incline is gSin70

Writing equations of motion,
At the highest point in the motion of projectile along the incline, the final vertical velocity is 0,
Vf = Vi - gSin70 t
0 = VoSin50 - gSin70 t
t = VoSin50 / gSin70
total time of flight = T = 2 times t = 2 VoSin50 / gSin70 = T ....1 st

In the horizontal along the incline,
L = Vi T + 1/2 (gcos70) T²
= VoCos50 T + 1/2 gCos70 T²
= VoCos50 (2 VoSin50 / gSin70) + 1/2 gCos70 (2 VoSin50 / gSin70)² [ Substitute T from 1st ]
= 15 Cos50 ( 2 * 15 Sin50 / 9.8Sin70) + 1/2 9.8 Cos70 ( 2 * 15 Sin50 /9.8Sin70)²
= 15 Cos50 ( 2.4955 ) + 1/2 9.8 Cos70 ( 2.4955 )²
= 9.6418 ( 2.4955 ) + 1.6758 ( 2.4955 )²
L =  34.4978 m