A physics professor did daredevil stunts in his spare time. His last stunt was a

Question

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.

Figure 1

I found the answer for part 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?) = 17.8 m/s

I need help answering part B

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

Explanation / Answer

B)

speed at top point, vo = 17.8/2 = 8.9 m/s

vox = 8.9*cos(53) = 5.36 m/s

voy = 8.9*sin(53) = 7.1 m/s

let t is the time taken to fall down

now Apply,

-h = voy*t - 0.5*g*t^2

-100 = 7.1*t - 4.9*t^2

4.9*t^2 - 7.1*t - 100 = 0

on sloving the above equation

we get, t = 5.3 s

so, distance treavelled in this time in horizonatl direction,

x = vox*t

= 5.36*5.3

= 28.4 m <<<<<<<<<<-------Answer

so he lands 28 m from the first bank.

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