1. Chanda, a stuntwoman in Hollywood, is preparing to use her motorcycle jump ov

Question

1. Chanda, a stuntwoman in Hollywood, is preparing to use her motorcycle jump over a burning bus from a ramp. The ramp is built so it will launch her at an anlge of 40 degrees and she needs to reach a maximum height of 10.0m above the top of the ramp in order to avoid catching fire from the bus. What does her speed need to be when she leaves the ramp? Please answer parts a, b, and c. Please show work how answers were found.

2. In order to perform this stund, Chanda must figure out how long the ramp needs ot be. She know that at the top of the ramp she needs to to reach the velocity (Vtop), which was solved in question1. She also knows that her motorcycle caused her to accelerate at 10m/s^2 as she goes up the ramp. Please answer parts a and b. Please show work how answers were found.

1. Chanda, a stuntwoman in Hollywood, is preparing to use her motorcycle jump over a burning bus from a ramp. The ramp is built so that it will launch her at an angle of 40% and she needs to reach a maximum height of 10.0 m above the top of the ramp in order to avoid catching fire from the bus. What does her speed need to be when she leaves the ramp a. Solve for the speed, Vtop, that she will need to leave the ramp with to safely land on the other side of the bus b. How far away does the landing ramp (assume she lands at the same height she begins) need to be placed? c. The director changes his mind at the last minute and decides that Chanda needs to land on top of a parked car that is 2.0 m lower than the top of the starting ramp. If she leaves the ramp with the same velocity, Vtop, how far away from the ramp does the car need to be so that she can land safely on the roof? 2 m Distance to car ? d. How much more time does it take Chandra to land on the parked car in c. than on the ramp in a

Explanation / Answer

vertical component of the velocity

vy  = vSin(40), v is the initial velocity on leaving the ramp.

Maximum height that can be reached is governed by

vy = sqrt(2gh) we have h = 10

vy = sqrt(2*9.8*10) = 14 m/s

launch velocity v = 14/Sin(40) = 21.78 m/s

maximum height is reached at the midpoint of the range.

Range R = v2Sin(2)/g = 21.782*Sin(80)/9.8

               = 47.67 m

the landing ramp needs to be at a distance of 47.67 m, so that she will land at the same height as that of launch.

vertical velocity when she reached the height of the launch will be same as that what she left = 14m/s

time to fall further 2 m is given by

2= 14*t+9.8t2/2

solving the above we get t= 0.136s

horizontal vel vx = 21.78Cos(40) = 16.68 m/s

horizontal distance traveled during the fall of 2 m

              = 16.68*0.136 = 2.269 m

The car needs to be placed at 47.67+2.27 = 49.94 m from the launching ramp.

        d) time t = 0.136 s as already calculated above.

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