A physics professor did dare - devil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. P3.63). The takeoff ramp was inclined at 53.0°, the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. Ignore air resistance.

(a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank?

(b) If his speed was only half the value found in part (a), where did he land?

Question

Physics

Zaria Bridges

16 December, 18:31

A physics professor did dare - devil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. P3.63). The takeoff ramp was inclined at 53.0°, the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. Ignore air resistance.

(a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank?

(b) If his speed was only half the value found in part (a), where did he land?

+3

Answers (1)

Know the Answer?

Jazmine Stein · Answer 1

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°, sinA = 0.8, cosA = 0.6

d = 40.0 m, h = 15.0 m, H = 100 m,

(a) find speed v

vertical displacement - h = v*sinA*t - 0.5gt^2

-15 = 0.8vt - 4.9t^2 ... (1)

horizontal displacement d = v*cosA*t = 0.6*v*t

so v*t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8*40/0.6 - 4.9t^2

t = 3.734 s

v = (40/0.6) / t = 17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = - H = v*sinA*t - gt^2/2

4.9*t^2 - 8.9*0.8*t - 100 = 0

t = 5.30 s

d = v*cosA*t = 28.3 m