How Fast Should the Professor Ride to Make the Jump?

In summary, the conversation discusses a physics professor who attempted a daredevil stunt of jumping over a river on a motorcycle. The professor had to consider the takeoff ramp's angle, the width of the river, and the height difference between the ramp and the far bank. The conversation also includes a formula for calculating the speed needed to reach the far bank and discusses what could go wrong with the stunt.
  • #1
Firben
145
0
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 ?
 
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  • #2
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.
 
  • #3
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 ?
 
  • #4
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.
 
  • #5



It is likely that there was an error in the calculations or the formula used. It is important to double check the values and units used in the calculations to ensure accuracy. It is also possible that the problem was set up incorrectly, in which case the solution would not be feasible. It is important to carefully read and understand the problem before attempting to solve it. Additionally, using a calculator or computer program with built-in formulas and units can help to reduce errors in calculations.
 

FAQ: How Fast Should the Professor Ride to Make the Jump?

What is projectile motion?

Projectile motion is the movement of an object through the air, where only the force of gravity acts upon it. This results in a curved path known as a parabola.

How is projectile motion related to jumping?

When a person jumps, their body becomes a projectile as it moves through the air. The force of gravity pulls the body down while the muscles provide an upward force, resulting in a parabolic trajectory.

What factors affect the projectile motion of a jump?

The factors that affect projectile motion of a jump include the initial velocity, angle of launch, air resistance, and the force of gravity.

How can the maximum height and range of a jump be calculated?

The maximum height and range of a jump can be calculated using the equations of motion and considering the initial velocity and angle of launch. These calculations can also be done using computer simulations.

How can the motion of a jump be optimized for maximum height or range?

The motion of a jump can be optimized by adjusting the angle of launch and the initial velocity. A shallower angle and higher initial velocity will result in a longer range, while a steeper angle and lower initial velocity will result in a higher maximum height.

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