3.67 Leaping the River II. A physics professor did daredevil stunts in his spare

Question

3.67 Leaping the River II. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. P3.67). The takeoff ramp was inclined a| 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 elow the ramp. You can 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?

Explanation / Answer

here,

theta = 53 degree

width of river , w = 40 m

height , y = - 15 m

a)

let the intial speed be u and the time taken be t

y = u * sin(theta) * t - 0.5 * g * t^2

- 15 = u * sin(53) * t - 0.5 * 9.81 * t^2 ....(1)

and

for horizontal direction

w = u * cos(theta) * t

40 = u * cos(53) * t ....(2)

from (1) and (2)

u = 17.9 m/s

the intial speed is 17.9 m/s

b)

the depth , y = - (15 + 100) m

intial speed , u' = u/2 = 8.95 m/s

let the time taken be t

y = u * sin(theta) * t - 0.5 * g * t^2

- 115 = 8.95 * sin(53) * t - 0.5 * 9.81 * t^2

t = 5.62 s

the horizontal distance , x = u * cos(theta) * t

x = 8.95 * cos(53) * 5.62  

x = 30.3 m

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.