Projectile Motion of a car off a cliff

[Vaalron]

I have a homework assignment, teacher didn't have corrections sheets so I'm checking my answers on here. Hope someone can help me out on some problems as well.


1) A diver runs horizontally wit a speed of 1.20 m/s off a platform that is 10.0m above the water.
(a) How long will it take the diver to hit the water?
I got 11.7s

I got this by dividing the horizontal distance by 1.2 m/s, getting my answer

(b) How far horizontally will the diver travel in this time?
I got 14.05m

I got this by using the formula xf^2= 1.2^2+2(9.8)(10)
vf^2=197.44
square it
get 14.05m



2)
A car drives straight off the edge of a vertical cliff that is 54m high. THe police at the scene of the accident note that the point of impact is 130m from the base of the cliff.
(a) how long was the car in the air?
I got 6.63, by doing 130/9.8

(b) what was the horizxontal speed of the car?
I got 64.974 m/s, by using the formula vf= vo+at^2



3)
The punter on a football team tries to kick a football so it stays in the air for a long "hang time." Suppose the ball is kicked with an inital velocity of 25.0m/s at an angle of 60.0 above the ground.

(a) What are the horizontal and vertical components of this velocity?

I got 21.65m for the vertical, 12.5 for horizontal or
y=21.65 and x=12.5
by doing sin60=y/25
and cos60=x/25

(b) How long will the football be in the air?

I can't figure this out, I put the variables in as this:

x=25.0 m/s
a=9.8 m/s
t=?
Vo=0
vf crossed out


Please help me?

(c) How far horizontally will the ball travel beore it hits the ground downfield?

12.5m


4)
A volleyball is spiked so that is has an initial velocity of 15 m/s directed downward at an angle of 55 below the horizontal. What is the horizontal component oof the ball's velocity when the opposing player fields the ball?

x=8.6

cos55=x/15 is what i did.



5)
A rock climber throws a small first aid kit to another climber who is higher up in the mountain. The initial velocity of the kit is 11 m/s at an angle of 65 above the horizontal. at the instant when the kit is caught, it is traveling horizontally, so it's vertical speed is zero. What is thevertical height between the two climbers?

sin65=x/11

x=9.97


6)
A motorcycle daredevil is attempting to jump across as many buses as possible, as shown below. The takeoff ramp makes an angle of 18.0 above the horizontal, and the landing ramp is identical to the takeoff ramp. the buses are parked side by side , and each bus is 2.74m wide. The daredevil leaves the ramp with a speed of 33.5 m/s. What is the max number of buses over whic hthe daedevil can jump?

I got 11.6 buses.

sin18= x/33.5
cos18=x/33.5


7)
A horizontal rifle is fired at a bulls eye. Tje muzzle speed is 670m/s. The barrel is pointed directly at the center of the bulls eye, but the bullet strikes the target 0.025m below the center. What is the horizontal distance between the end of the rifle and the bulls-eye?

I got 1.34, by doing .025/9.8

**8**)
A rocket is fired at a speed of 75.0m/s from the ground level, at an angle of 60.0 above the horizontal. The rocket is fired toward an 11.0m wall, which is located 27.0m away. By how much does the rocket clear the top of the wall?

I couldn't do this, Here's a figure that I thought would work, but don't know where to go from there:


http://img388.imageshack.us/img388/2676/blahae8.jpg
**EDIT** Forgot to put the 60 degrees on the figure, sorry.

Sorry for all the questions, It's due tomarrow and I have no corrections sheet. I don't really understand this much at all, nor any other part of physics, So I'm in need of someone good at physics to check all these, :/

 

[hotvette]

Double post. Replies are in the other one:

https://www.physicsforums.com/showthread.php?t=188807

 

[m2003]

Hello,

I am happy to help you with your homework assignment. First, I would like to commend you for taking the time to check your answers and seeking help when needed. This shows a dedication to learning and understanding the material.

Now, let's go through your answers and see if they are correct.

1) Your answer for part (a) is correct. The time it takes for the diver to hit the water is 11.7 seconds.

For part (b), you are using the correct formula, but you made a mistake in your calculation. The correct answer is 14.04 meters. This is obtained by using the formula xf = x0 + v0t + 1/2at^2, where xf is the final position, x0 is the initial position (which is 0 in this case), v0 is the initial velocity (which is 1.2 m/s), t is the time (which is 11.7 seconds), and a is the acceleration (which is 0 since there is no acceleration in the horizontal direction).

2) Your answer for part (a) is correct. The car was in the air for 6.63 seconds.

For part (b), you used the correct formula, but you made a mistake in your calculation. The correct answer is 66.97 m/s. This is obtained by using the formula vf = vo + at, where vf is the final velocity (which is 0 since the car hits the ground and stops), vo is the initial velocity (which is unknown), a is the acceleration (which is -9.8 m/s^2 since the car is falling), and t is the time (which is 6.63 seconds).

3) Your answers for part (a) are correct. The horizontal component of the velocity is 21.65 m/s and the vertical component is 12.5 m/s.

For part (b), you can use the formula y = y0 + v0yt + 1/2at^2 to find the time the football is in the air. The initial vertical position (y0) is 0, the initial vertical velocity (v0y) is 12.5 m/s, and the acceleration in the vertical direction (a) is -9.8 m/s^2. Solving for t, we get t = 2.55 seconds. This means the football will be in the air for