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 takeoff jump was inclined at 53.0 degrees, the river was 40.0 m wide, and the far bank was 15.0m lower than the top of the ramp. The river itself was 100 m below 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 (a), where did he land?

Question

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Manetho Manetho · Answer 1 · 2019-12-23T09:50:20+01:00

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

[tex]-h= vsinA\times t-gt^2/2[/tex]

putting values h=15 m, v=0.8

[tex]-15 = 0.8vt - 4.9t^2[/tex] ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

[tex]-15 = 0.8\times40/0.6 - 4.9t^2[/tex]

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 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 = [tex]-H =v sinA t - gt^2/2[/tex]

⇒ [tex]4.9t^2 - 8.9\times0.8t - 100 = 0[/tex]

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m