A ball is launched with a velocity of magnitude 10.0 m/s, at an angle of 50.0° to the horizontal. The launch point is at the base of a ramp of horizontal length d1 6.00 m and height d2 3.60 m. A plateau is located at the top of the ramp. Required:
a. Does the ball land on the ramp or the plateau? When it lands.
b. What are the magnitude and angle of its displacement from the launch point?

Kindly check explanation

Explanation:

Given the following :

Initial Velocity (u) = 10m/s

Angle (θ) = 50°

Horizontal length d1 = 6m

Height d2 = 3.6m

g = acceleration due to gravity = 9.8m/s²

Horizontal component :

X = U× cosΘ = 10 × cos 50 = 10 × 0.6429 = 6.43t

X = 6.43t

Vertical component :

Y = Usinθ - 1/2gt² = 10 ×sin 50 - 1/2(9.8)t² = 7.66t - 4.9t²

From X :

t = X / 6.43

Hence,

Y = 7.66(X/6.43) - 4.9(X/6.43)²

Y = 1.19X - 0.12X²

Slope of ramp Y':

(3.6 ÷ 6)X = 0.6X

Hence,

Y = Y'

1.19X - 0.12X² = 0.6X

-0.12X² = 0.6X - 1.19X

-0.12X² = - 0.59X

X² / X = 0.59 / 0.12

X = 4.92 m

Ball hits the ramp as X is < 6m

b.

explanation:

10.0

c

explanation: