Problem: A physics professor did daredevil stunts in his spare time. His last st

Question

Problem: A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (the figure (Figure 1) ). 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. You can ignore air resistance.

Question:

1.What should his speed have been at the top of the ramp to have just made it to the edge of the far bank?

2.If his speed was only half the value found in A, where did he land?

Explanation / Answer

@ , angle = 53 degree

Let d = 40 m

h= 15m

H= 100m

vertical distance , Sy = uy*t + (1/2)*a*t2

[ uy = V*sin 53, a(acceleration)= -g , t = time]

  Sy = uy*t + (1/2)*a*t2

   -h = V*sin 53 *t - (1/2)*g*t2

-15 = 0.798*V*t - 4.9t2 ---------------------------1 ( g= 9.8m/s2 )

horizontal distance , Sx= ux*t + (1/2)*ax*t2

[ here ax= horizontal acceleration = 0, Sx=d ,ux = horizontal velocity = V*cos 53]

d= V*cos 53*t + 0

40 = 0.6018*V*t

Vt = 40/ 0.6018 = 66.467

by putting this value in eq 1:

-15 = 0.798*V*t - 4.9t2

-15 = 0.798 *66.467 - 4.9*t2

-68.04 = -4.9*t2

t= 3.7 s

Since Vt = 66.467

V( speed) = 66.467 / 3.7 = 17.96 m/s

------------------------------------------------------

b) now V= 17.96 / 2 = 8.98 m/s

d = where did he land = ?

vertical distance , - H = V*sin 53*t - (1/2)*g*t2

-100 = 8.98*sin53*t - (1/2)*9.8*t2

-100 = 7.17*t - 4.9*t2

4.9*t2 - 7.17 t -100 = 0

on solving this equation, t= 5.3 s

horizontal distance , d= V*cos53*t

= 8.98*cos53*5.3 = 28.64 m

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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