Aball is thrown with initial speed v0 up an inclined plane. the plane is inclined at an angle φ above the horizontal, and the ball’s initial velocity is at an angle θ above the plane. choose axes with x measured up the slope and y normal to the slope. you can neglect the z coordinate as the ball remains in the xy-plane (z = 0).

(a) write down newton’s second law using these axes and find the ball’s position as a function of time. (your solution only needs to be valid for times before the ball lands back on the inclined plane.)

(b) show that the ball lands on the plane a distance r = 2v0 2 sinθ cos(θ + φ) / (gcos2 φ) from its launch point.

(c) show that for given values of v0 and φ, the maximum possible range up the inclined plane is rmax = v0 2 /[g(1+ sinφ)].

Explanation:

a )  The plane is making angle φwith horizontal and ball is thrown with velocity v₀  at angle θ with the inclined plane . In this situation the only thing that should be taken care is that , instead of g , the component of g that is g cos φ will act on the object .

If t be the time after which ball reaches position y with respect to inclined plane

initial velocity with respect to inclined plane

u = v₀ sinθ vertically upwards with respect to inclined plane

Apply the formula

s = ut -1/2 gt²

y =  v₀ sinθ t - 1/2 gcosφ t²

b )

Displacement along the inclined plane R

Time of flight T

Use the formula

v = u - gt

0 = v₀ sinθ - gcosφ t

t = v₀ sinθ / gcosφ

Time of flight

T = 2t = 2 x v₀ sinθ / gcosφ

R = v₀ cosθ X T - 1/2  gsinφ x ( T)²

= 2v₀ cosθ x v₀ sinθ / gcosφ - 2  gsinφ x v₀² sin²θ / g²cos²φ

= 2v₀ ² sinθ (  cosθ / gcosφ - sinθsinφ /  gcos²φ )

= 2v₀ ² sinθ (cosθcosφ - sinθsinφ ) / gcos²φ

R = 2v₀ ²  sinθ cos(θ+ φ ) / gcos²φ

568.98

explanation:

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