How Does Projectile Angle Affect Flight Time, Max Height, and Initial Speed?

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In summary, the initial speed of the ball thrown straight up is 29.43 m/s. The maximum height is determined by the time of flight, which is 3 seconds. When the ball is thrown at an angle other than 90 degrees, the time of flight may change, but the maximum height is still determined by the total flight time. The initial speed can also be determined, assuming it is thrown at the same speed as the ball thrown straight up.
  • #1
sbright
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Homework Statement


Find initial speed and maximum height of a ball that is thrown 90 degrees straight up and lands on the ground in 3 seconds. What happens when the ball is thrown at some angle other than 90 degrees? Does it change the time of flight? Can you still determine the max height and initial speed?


Homework Equations


v=vo+at
Vy=Vsin90
y=vot+1/2gt^2


The Attempt at a Solution


For initial speed i did 9.81*3=29.43 to find velocity then (29.43)sin90=29.43 which i believe is the initial speed(Vo)
I'm having trouble with max height.
(29.43)(3)+4.9(3^2)=132.39m
Which i believe is much to big of a number..help please
Also I think as you move closer to 0 degrees and further away from 90 degrees, time of flight decreases, but can we determine max height and initial speed?
 
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  • #2
sbright said:

Homework Statement


Find initial speed and maximum height of a ball that is thrown 90 degrees straight up and lands on the ground in 3 seconds. What happens when the ball is thrown at some angle other than 90 degrees? Does it change the time of flight? Can you still determine the max height and initial speed?


Homework Equations


v=vo+at
Vy=Vsin90
y=vot+1/2gt^2


The Attempt at a Solution


For initial speed i did 9.81*3=29.43 to find velocity then (29.43)sin90=29.43 which i believe is the initial speed(Vo)
I'm having trouble with max height.
(29.43)(3)+4.9(3^2)=132.39m
Which i believe is much to big of a number..help please
Also I think as you move closer to 0 degrees and further away from 90 degrees, time of flight decreases, but can we determine max height and initial speed?

Firstly, this ball only traveled up for 1.5 seconds, then down for 1.5 seconds which is why it landed after 3 seconds.
 
  • #3
PeterO said:
Firstly, this ball only traveled up for 1.5 seconds, then down for 1.5 seconds which is why it landed after 3 seconds.

So would this be correct
29.43(1.5)+4.9(1.5^2)=55.17
still seems like too big of a number...
 
  • #4
Ok so I'm pretty sure I found the right answer
29.43(1.5)-4.9(1.5^2)=33.12m

Now if anyone could help with the very last question, "What happens when you throw a ball at some angle other than 90 degrees? Can you still determine the maximum height and initial speed? Give careful explanations in your answers.

Any input with that last question would make me the happiest person in the world thank you!
 
  • #5
Remember* only given time (for that last question)
 
  • #6
sbright said:
Ok so I'm pretty sure I found the right answer
29.43(1.5)-4.9(1.5^2)=33.12m

Now if anyone could help with the very last question, "What happens when you throw a ball at some angle other than 90 degrees? Can you still determine the maximum height and initial speed? Give careful explanations in your answers.

Any input with that last question would make me the happiest person in the world thank you!

I find the follow up a little ambiguous.

What happens when the ball is thrown at some angle other than 90 degrees? Does it change the time of flight? Can you still determine the max height and initial speed?


You are asked "does it change the time of flight?"
Well if the ball still lands 3 seconds later, clearly no. However if we are to assume that it was thrown at the same speed you calculate in the first part, then clearly yes.

The maximum height is always determined by the flight time.
If an object is in the air for 3 seconds, it travels up for 1.5 seconds and down for 1.5 seconds. How far down does anything fall in 1.5 seconds [always ignoring iar resistance]
 
  • #7
sbright said:
Ok so I'm pretty sure I found the right answer
29.43(1.5)-4.9(1.5^2)=33.12m

Now if anyone could help with the very last question, "What happens when you throw a ball at some angle other than 90 degrees? Can you still determine the maximum height and initial speed? Give careful explanations in your answers.

Any input with that last question would make me the happiest person in the world thank you!

That answer is far too high, since your initial velocity was based on 3 seconds to each maximum height, and it is only 1.5 seconds.
 

FAQ: How Does Projectile Angle Affect Flight Time, Max Height, and Initial Speed?

What is the equation for calculating the height of a ball thrown upward given time?

The equation for calculating the height of a ball thrown upward given time is h(t) = -16t^2 + v0t + h0, where h(t) is the height of the ball at time t, v0 is the initial velocity, and h0 is the initial height.

What is the significance of the constant -16 in the equation?

The constant -16 in the equation represents the acceleration due to gravity, which is -9.8 meters per second squared. This value is negative because it acts in the opposite direction of the ball's motion.

How can I determine the maximum height reached by the ball?

The maximum height reached by the ball can be determined by finding the vertex of the parabolic trajectory, which is the highest point. This can be done by finding the time at which the velocity is equal to zero, and then plugging that time into the equation for height.

Is air resistance taken into account in this equation?

No, this equation does not take into account air resistance. Air resistance can significantly affect the trajectory of a thrown ball, especially at higher velocities.

Can this equation be used to calculate the time taken for the ball to reach a certain height?

Yes, this equation can be rearranged to solve for time (t) when the height (h) is known. However, it assumes that the ball is thrown from a certain initial height and does not take into account any obstacles or changes in velocity during the trajectory.

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