shoots off a ramp with a velocity of 6.8 m/s, di be parallel to the ground, the

Question

shoots off a ramp with a velocity of 6.8 m/s, di be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. The end of the ramp is 1.5 m above the ground. Let the x axis (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp? Section 3.3 A car drives straight off the edge of a cliff that is 57 m high. The police at the scene of the accident observed that the point of impact is 122 m from the base of the cilf. How 28

Explanation / Answer

u = 6.8 m/s
angle = 63

h = 1.5 m

(a)
v^2 = u^2 + 2as
0 = (6.8*sin(63))^2 - 2*9.8*s
s = 1.873 m

s = 1.873 + 1.5 = 3.373 m

(b)
Let the time be t
v = u + at
0 = (6.8*sin(63)) - 9.8*t
t = 0.62 sec

Horizontal Distance = 0.62*(6.8*cos(63)) = 1.914 m

Please upvote if the answers are correct.

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