Aphysics 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?

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

putting values h=15 m, v=0.8

 ............. (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

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 =

⇒

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.