Double Check My Answers: Calculating Projectile Motion with Initial Velocity

[silentcoder]

Hi,
I am not going to ask you to do my homework, since i have done them myself. I just want your help to double check my questions.

1. A tennis ball is thrown horizontally with an initial velocity of 14.5 m/s from a height of 1.4 m above the ground. Determine the following:

(a) the time the ball spends in the air. (0.53 s)

(b) the velocity with which the ball impacts the ground. (15 m/s [20 degree below horizonal])

2. You are trying to throw a ball over a 3.0 m high fence. You throw the ball from a height of 1.0 m with an initial velocity of 27 m/s 50° above the horizontal. Determine the following:

(a) the maximum height of the ball. (23.5 m)

(b) the time the ball spends in the air. (4.33 s)

3. An artillery gun can fire a shell at an initial velocity of 1700 m/s at an angle of 55° above the horizontal. Determine the following:

(a) the time the shell spends in the air. (Is the answer appx. 284 s)

If i am missing any directions, please let me know.
Thanks! I really appreciate it!

 

[SammyS]

Hi,
I am not going to ask you to do my homework, since i have done them myself. I just want your help to double check my questions.

1. A tennis ball is thrown horizontally with an initial velocity of 14.5 m/s from a height of 1.4 m above the ground. Determine the following:

(a) the time the ball spends in the air. (0.53 s)

(b) the velocity with which the ball impacts the ground. (15 m/s [20 degree below horizonal])

2. You are trying to throw a ball over a 3.0 m high fence. You throw the ball from a height of 1.0 m with an initial velocity of 27 m/s 50° above the horizontal. Determine the following:

(a) the maximum height of the ball. (23.5 m)

(b) the time the ball spends in the air. (4.33 s)

3. An artillery gun can fire a shell at an initial velocity of 1700 m/s at an angle of 55° above the horizontal. Determine the following:

(a) the time the shell spends in the air. (Is the answer appx. 284 s)

If i am missing any directions, please let me know.
Thanks! I really appreciate it!

Hello silentcoder. Welcome to PF !

It makes much more sense for us to check your answers, if you at least give some detail as to how you got your answers. Often showing your work is even better.

You will tend to get better responses if you do these things.​

1. a) is correct to 2 sig. fig.

1. b): I get about the same angle -- not the same magnitude.

I'll look at the others & get back.

 

[silentcoder]

Thanks, next time i will post my solutions as well.

For 1. (b) I get -5.194 as my final velocity using equation: vf = vi + 2at
Wouldn't it make sense if you get same angel, then shouldn't be the resultant velocity be same as well?

 

[SammyS]

Thanks, next time i will post my solutions as well.

For 1. (b) I get -5.194 as my final velocity using equation: vf = vi + 2at
Wouldn't it make sense if you get same angel, then shouldn't be the resultant velocity be same as well?

Yes, I agree with that statement.

I don't agree with 15 m/s .

 

[silentcoder]

Yes, I agree with that statement.

I don't agree with 15 m/s .

What value are you getting for 1.(b)?
and is the rest of question 2 correct?

 

[haruspex]

All of your answers seem to be about right, just a little inaccurate. For 1b I get 15.4 m/s; for 2a, 22.8m; 2b, 4.26s; ... If improving your accuracy doesn't fix these, pls post your detailed working.

 

[SammyS]

Thanks, next time i will post my solutions as well.

For 1. (b) I get -5.194 as my final velocity using equation: vf = vi + 2at
Wouldn't it make sense if you get same angel, then shouldn't be the resultant velocity be same as well?

It looks like the time you used for that was the rounded value you listed as the answer for 1 a .

Keep a couple of extra digits for anything you use as an intermediate result to be used in a further calculation.

I get about -5.23832 m/s for the vertical component of the velocity at impact.

θ ≈ tan^-1(-5.23832/14.5) ≈ -20°

What value are you getting for 1.(b)?
and is the rest of question 2 correct?

I get answers that are close to what you did for #2, but the difference is too much to be chalked up to round-off error.

How did you get them?

 

[silentcoder]

for 2.(a) I did:

vf^2 = Vi^2 + 2ad
         0 m/s = (21 m/s)^2 + s(-9.8 m/s^2)d
         0 m/s = (441 m/s) - (19.6 m/s^2) d
         0 - 441 m/s = -19.6 m/s^2 d
          -441m/s / -19.6 m/s^2 = d
         d = 22.5 m
         since ur throwing the ball from a height of 1m: [COLOR="red"]22.5m + 1m = 23.5 m[/COLOR]

for 2.(b) I did:

     d = vi t + 1/2 at^2
     -1m = (21 m/s)t - 1/2(-9.8 m/s^2) t^2
     -1m = (21 m/s)t - 4.9 m/s^2 t^2
     = -4.9 m/s^2 t^2 + 21 m/s t + 1m
Quadratic equation: t = -0.05 and [COLOR="Red"]4.33s[/COLOR]

 

[haruspex]

for 2.(a) I did:

vf^2 = Vi^2 + 2ad
         0 m/s = (21 m/s)^2 + s(-9.8 m/s^2)d

But 21 is not very accurate for 27 sin (50^o). It's about 20.7.

 

[SammyS]

for 2.(a) I did:

         vf^2 = Vi^2 + 2ad
         0 m/s = (21 m/s)^2 + s(-9.8 m/s^2)d
         0 m/s = (441 m/s) - (19.6 m/s^2) d
         0 - 441 m/s = -19.6 m/s^2 d
          -441m/s / -19.6 m/s^2 = d
         d = 22.5 m
         since ur throwing the ball from a height of 1m: [COLOR="red"]22.5m + 1m = 23.5 m[/COLOR]

for 2.(b) I did:

     d = vi t + 1/2 at^2
     -1m = (21 m/s)t - 1/2(-9.8 m/s^2) t^2
     -1m = (21 m/s)t - 4.9 m/s^2 t^2
     = -4.9 m/s^2 t^2 + 21 m/s t + 1m
Quadratic equation: t = -0.05 and [COLOR="Red"]4.33s[/COLOR]

As I pointed out earlier, you should retain a few digits beyond the rules for significant digits when using intermediate results.

(27)sin(55°) ≈ 20.68
In my opinion, it's best to work as much of the problem symbolically as possible.

Solving the following for d:

$\displaystyle {v_f}^2 = {v_i}^2 + 2ad$

gives:

$\displaystyle d=\frac{{v_f}^2-{v_i}^2}{2a}$

Substituting the appropriate values gives:

$\displaystyle d=\frac{{0}^2-(27\,\sin(55^\circ))^2}{2(-9.8)}$

This is approximately 24.9575 .