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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?

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Answer:

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

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