Calculating Displacement and Landing Position of a Horizontal Projectile

[warrior2014]

Homework Statement

A tennis ball is thrown horizontally from an elevation of 13.03m above the ground with a speed of 24.0m/s.

a)where is the ball after 1.50 seconds (both horizontally and vertically)?

b) if the ball is still in the air, how long before it hits the ground?

c) if the ball is still in the air, where will it be with respect to the starting point once it lands?

I'd really appreciate it if you'd be able to help me as soon as possible. Thanks!

The Attempt at a Solution

for the first part, I use the displacement equation and found the horizontal displacement to be 36m and the vertical displacement to be 2.005m. For part b and c, I'm not sure how to start the question.

 

[Saitama]

For simplicity, let's make a table of a given data.

H-Horizontal Motion
V-Vertical Motion
s_H-Displacement in horizontal direction
s_V-Displacement in vertical direction
u-initial velocity
t-time

Okay, for the first part, how you get your displacement to be 2.005m for the vertical motion?
Show some steps.

 

[warrior2014]

I used the equation d=Vy(t) + 0.5(ay)(t^2)
=(0)(1.5) + (0.5) (-9.8)(1.5^2)
=2.005m

 

[rl.bhat]

Check the calculations.
d = (0.5)(9.8)(2.25) = ..?

 

[warrior2014]

you get -11.025m from the displacement equation and add 13.03m because its that much higher than the ground. I have confirmed that this answer is correct but am unsure how to do part b and c.

 

[Saitama]

For part b, you can do this:-
We have to calculate the time when it reaches the ground.
For that you can have 13.03m in the displacement equation and find the time.

For part c:-
When you get the time for vertical motion from part b, it would be same for horizontal motion.

Try solving.

 

[warrior2014]

if the ball is 36m horizontal and 2.005m vertical at 1.5 seconds, for part b if I set d=13.03 and use v=24m/s, then t=0.54s. This doesn't make sense because at 1.5 seconds its still in the air so how can it hit the ground at 0.54s?

 

[Saitama]

if the ball is 36m horizontal and 2.005m vertical at 1.5 seconds, for part b if I set d=13.03 and use v=24m/s, then t=0.54s. This doesn't make sense because at 1.5 seconds its still in the air so how can it hit the ground at 0.54s?

2.005m is the height above the ground at 1.5 seconds.
You used initial velocity as 24m/s for vertical motion, which is wrong. The initial velocity for vertical motion is 0m/s. Check the table above. :)

 

[warrior2014]

okay so for part b, I put d=13.03 v=0, a=9.8 and used the displacement equation to find time which gave me 1.63s...but this was wrong. But for part c, when I used 1.63s, it gave me the write displacement answer. I am unsure as to what I did wrong with part b.

 

[Saitama]

okay so for part b, I put d=13.03 v=0, a=9.8 and used the displacement equation to find time which gave me 1.63s...but this was wrong. But for part c, when I used 1.63s, it gave me the write displacement answer. I am unsure as to what I did wrong with part b.

What's the answer then, in case if you have the answers?

 

[warrior2014]

I don't have the answers- it is an online question so when I get it wrong, it just says wrong. would you have any other suggestions??

 

[Saitama]

I don't have the answers- it is an online question so when I get it wrong, it just says wrong. would you have any other suggestions??

Are there any options, if so please post them here. :)

 

[warrior2014]

no there aren't any options! sorry, but would you be able to help me with the following question?:

A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 32 m/s at an angle of 50° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.

a) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land?

For this question, I used the equation for Range and got 102.9m but it was incorrect.

b)What is the y-component of the cannonball's velocity just before it lands? The y-axis points up.

For this question, I did Vy=32sin50=24.5 (but this was also incorrect). Any help will be appreciated!

 

[rl.bhat]

okay so for part b, I put d=13.03 v=0, a=9.8 and used the displacement equation to find time which gave me 1.63s...but this was wrong. But for part c, when I used 1.63s, it gave me the write displacement answer. I am unsure as to what I did wrong with part b.

1.63s - 1.5s = .? is the required answer. Check it.
For the second problem, try the following formula.
-y = x*tanθ - 0.5*g*x^2/(v^2*cos^2θ) and solve for x.
For part b, the ball is landing bellow the point of projection. Find time of flight and use
-vy = -v*sinθ - gt. to get vy.

 

[warrior2014]

yes that was the correct answer. thank you!