Projectile being launched from ground level

[AryRezvani]

Homework Statement

A projectile is fired from ground level with an initial velocity of 50 m/s and an initial angle of 37°. Assuming g = 9.8 m/s^2, find:

a) the projectiles total time of flight.
b) the maximum height attained
c) the total horizontal distance traveled
d) the final horizontal and vertical velocities just before it hit the ground.

Homework Equations

okay, so beginning with a), I believe the necessary equation is h=v_i²Sin²θ_i/2g

The Attempt at a Solution

Plug in a) into my calculator, and I get uhh 4436.721196. I'm not sure if I'm using the correct method.

 

[Saitama]

okay, so beginning with a), I believe the necessary equation is h=v_i²Sin²θ_i/2g

You need to find the total time of flight in a), the equation you posted is for the max. height attained by the projectile.

 

[AryRezvani]

You need to find the total time of flight in a), the equation you posted is for the max. height attained by the projectile.

would you utilize V_f=V_i+at?

So V_f = 0 (at the top of the trajectory V_f would be zero.
g = -9.8
V_i = 50

Plus in so you get -9.8t = -50
t = 5.102?

 

[Saitama]

would you utilize V_f=V_i+at?

So V_f = 0 (at the top of the trajectory V_f would be zero.
g = -9.8
V_i = 50

Plus in so you get -9.8t = -50
t = 5.102?

For the vertical motion, what's the initial velocity? You need to take the component of initial velocity in the vertical direction for V_i.

 

[AryRezvani]

For the vertical motion, what's the initial velocity? You need to take the component of initial velocity in the vertical direction for V_i.

Hmm I got 50*Sin(37) which gave me 30.09075116.

 

[Saitama]

Hmm I got 50*Sin(37) which gave me 30.09075116.

You shouldn't get such a big answer. The component you took is correct. Check your calculations. Maybe, it will help if you show us your steps. Make sure you double the time you get when you plug the values in the equation V_f=V_i+at because you need to find the total time of flight. The projectile reaches the max. height and comes back again.

 

[AryRezvani]

Okay, well in my textbook, it says the following formula is utilized to find V_yi:

V_yi=V_iSin(θ_i)

I plugged in the numerical values into my calculator and got V_yi = 30.09075116.

Now, I plugged that answer into V_yf = V_yi + a_yt

V_yf should equal zero because at the top of the trajectory the y component is zero.

I solved and got t = 3.09075116 seconds.

Edit; multiply the answer by 2 and you get 6.14096... Look good?

 

[Saitama]

Okay, well in my textbook, it says the following formula is utilized to find V_yi:

V_yi=V_iSin(θ_i)

I plugged in the numerical values into my calculator and got V_yi = 30.09075116.

Oops, sorry, i thought you meant time of flight when you wrote 30.09075116.

Now, I plugged that answer into V_yf = V_yi + a_yt

V_yf should equal zero because at the top of the trajectory the y component is zero.

I solved and got t = 3.09075116 seconds.

Just double the time you calculated, for the reason i mentioned in my previous post.

EDIT: Just saw your edit, looks good.

 

[AryRezvani]

Oops, sorry, i thought you meant time of flight when you wrote 30.09075116.


Just double the time you calculated, for the reason i mentioned in my previous post.

EDIT: Just saw your edit, looks good.

Thanks :)

If you could, could you just quickly check my work for the rest of them?

In regards to b), I used the equation I mentioned initially (V_i²Sin²θ_i/2g

Plugged in (50^2)*Sin(37)*Sin(37)/2*9.8 and got 4436.721196.

Look a little off?

 

[Saitama]

Thanks :)

If you could, could you just quickly check my work for the rest of them?

In regards to b), I used the equation I mentioned initially (V_i²Sin²θ_i/2g

Plugged in (50^2)*Sin(37)*Sin(37)/2*9.8 and got 4436.721196.

Look a little off?

Its too much off.
Check your calculations again. Or show the steps.

 

[AryRezvani]

Its too much off.
Check your calculations again. Or show the steps.

Ah, yeah, incorrect calculations lol. I fixed it and got something around 46.196..

Okay, so part c) I found the V_xi to be 39.931775 or does this have something to do with range (R)?

 

[Saitama]

Okay, so part c) I found the V_xi to be 39.931775 or does this have something to do with range (R)?

Yes, you need to find the range.

 

[AryRezvani]

c) V_xi = 39.9317755

Formula for R being: V_xi2t_a

T_a being the halfway point of the trajectory so all in all this comes out to be the time we calculated earlier.

R = (39.9317755)(3.07048412)2

R = 245.2198204 meters

Look good?

 

[Saitama]

c) V_xi = 39.9317755

Formula for R being: V_xi2t_a

T_a being the halfway point of the trajectory so all in all this comes out to be the time we calculated earlier.

R = (39.9317755)(3.07048412)2

R = 245.2198204 meters

Look good?

Looks correct to me.

 

[AryRezvani]

Looks correct to me.

Okay, lastly, d).

Hmm okay, well horizontal velocity is constant so I don't think there is a way of doing the horizontal one.

Wouldn't these just be their respective initial velocities? Or maybe you'd plug them into an equation that has a V_f and solve.

V_f = V_i + at

Time being how long the ball is in the air.

 

[Saitama]

Okay, lastly, d).

Hmm okay, well horizontal velocity is constant so I don't think there is a way of doing the horizontal one.

Wouldn't these just be their respective initial velocities? Or maybe you'd plug them into an equation that has a V_f and solve.

V_f = V_i + at

Time being how long the ball is in the air.

Horizontal velocity will be the same, vertical velocity will be opposite in direction to that of initial. It asks about velocity, not speed.

 

[AryRezvani]

Horizontal velocity will be the same, vertical velocity will be opposite in direction to that of initial. It asks about velocity, not speed.

Understood, so the object fired has an initial velocity of +50 m/s; however, as it comes down, the velocity is the same, but simply decreasing?

So the velocity just as the object hits the ground is -50 m/s?

Edit: Leaving for class right now. Thanks for taking the time to explain this to me. I really appreciate the help. :)

 

[Saitama]

Understood, so the object fired has an initial velocity of +50 m/s; however, as it comes down, the velocity is the same, but simply decreasing?

So the velocity just as the object hits the ground is -50 m/s?

No, i did not mean that. The question asks about the vertical and horizontal final velocities. The direction of final vertical velocity would be the opposite to that of the initial vertical velocity. You cannot say that its -50 m/s.

 

[AryRezvani]

No, i did not mean that. The question asks about the vertical and horizontal final velocities. The direction of final vertical velocity would be the opposite to that of the initial vertical velocity. You cannot say that its -50 m/s.

So what would it be? Would it be 50 m/s simply on the opposite direction? In other words, 50 m/s with a downward direction?