Projectile motion of a jump

[Firben]

1. 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 ramp was inclined at 53 degrees, the river was 40 m. wide and the far bank was 15 m 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. Is his speed was only half the value found in (a), where did he land?



2.
x = vcos(alpha)*t
y = vtsin(alpha) - 1/2gt^2



3. 0 = vcos53 * t ==> t = 40/(vcos53)

1 in 2 ;

85 = v * 40/(vcos53) * sin53 - 1/2 *9.81 * (40/(vcos53))^2

But i got a imaginary number for v

Homework Statement

What went wrong ?

 

[Simon Bridge]

You correctly put the far bank 85m above the river but failed to account for the height of the ramp above the river (100m). Try putting the final height as -15m

These problems work better by drawing the v-t diagrams for the components.

 

[Firben]

Well i got the right answer at a) but in b) but if his speed only was half the value found in a) where sis he land ?

v0 = 8.9 ms, v = 53

x = v0cos(alpha)t

y = v0tsin(alpha) - 1/2g*t^2

What should i do next ?

 

[Simon Bridge]

Well - he's either going to splash in the river or splat against the cliff.
So = work out the distance if he should land at river-height. If this distance is more than the width of the river then he splat's the cliff right?

Formally, you can work out the trajectory y(x) compare with the horizontal line at river height and the vertical line at the cliff distance.