During a movie scene, a motorcycle is to drive off the roof of a 44 meter tall building and land 2.3 seconds later. If the motorcycle leaves the first roof traveling horizontally at 29.0 m/s, how far away from the building does the vehicle land and with what vertical velocity at impact?

1. 66.7 m

2. 51.54 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) = 44 m

Time (t) = 2.3 s

Initial velocity (u) = 29.0 m/s

Horizontal distance (s) =.?

Final Velocity (v) =.?

1. Determination of the horizontal distance travelled.

Time (t) = 2.3 s

Initial velocity (u) = 29.0 m/s

Horizontal distance (s) =.?

s = ut

s = 29 × 2.3

s = 66.7 m

Therefore, the vehicle will land 66.7 m from the building.

2. Determination of the velocity at impact.

Time (t) = 2.3 s

Initial velocity (u) = 29.0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Final Velocity (v) =.?

v = u + gt

v = 29 + (9.8 × 2.3)

v = 29 + 22.54

v = 51.54 m/s

Therefore, the vehicle hit the ground with a velocity of 51.54 m/s

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explanation:

i think its c if i'm wrong