Projectile, Range, Maximum Height, Time of Flight Problem

In summary, the ball takes 1.3854 seconds to reach its maximum height when thrown by the teacher with an initial velocity of 17 m/s at a 53 degree angle. To calculate the time it takes to reach the hoop, first find the maximum height and then use this height plus the difference between the starting and ending heights to calculate the total time. The horizontal distance of the shot can be found by using S = VT, where V is the initial velocity and T is the total time calculated.
  • #1
Gimp Mask
1
0

Homework Statement



Your teacher tosses a basketball. The ball gets through the hoop. How long does it take the ball to reach its maximum height? How long does it take the ball to reach the hoop? What is the horizontal length of the shot? (Neglect air friction).

Initial velocity: 17 m/s
53 degree angle between the ball's initial position and the ground.
The height of the teacher is 2.576 m. (This is the ball's initial height.)
The height of the goal is 3.048 m. (This is the ball's final height.)


Homework Equations



Vf = Vyo + at
Vyo = VoSinΘ



The Attempt at a Solution



I solved the first part as follows:
Vyo = VoSinΘ
Vf = Vyo + at
0 = 13.5768 - 9.8t
-13.5768 = -9.8t
t = 1.3854s (Time to get to maximum height. This answer is correct.)

I'm having some trouble with the second and third part though. For the second part:
Vyo = VoSinΘ
d = Vyot + 1/2at^2

I subtracted the two heights (3.048m - 2.576m = 0.472m).
0.472m = 13.5768t + 1/2(9.8m/s^2)t^2

I then used the quadratic formula:
-13.5768 +- (square root of: ((13.5768^2) - 4(4.9)(-0.472))/(2)/(-0.472)

The answer I derived was not correct (28.4).

For the third part:

Xmax = 2Vo^2sinΘcosΘ/g

2(17^2)(sin53)(cos53)/9.8

The answer I derived was not correct (28.3).

Please help me if you can!
 
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  • #2
For the second part, first calculate the time it takes the ball to reach maximum height, which you already found to be 1.4 seconds.

Next, calculate the maximum height it reached. From there, the object is falling the maximum height PLUS the .472 meter difference. Use that total height to calculate the total time. Because the object is simply falling from the maximum height, Vo is 0 and so it's not a quadratic equation.

Once you find this time, add that to the first time you got. This will give you the total time.

Once you have the total time, you can simply use S = VT to solve for the horizontal distance.

Hope that helps :)



Gimp Mask said:

Homework Statement



Your teacher tosses a basketball. The ball gets through the hoop. How long does it take the ball to reach its maximum height? How long does it take the ball to reach the hoop? What is the horizontal length of the shot? (Neglect air friction).

Initial velocity: 17 m/s
53 degree angle between the ball's initial position and the ground.
The height of the teacher is 2.576 m. (This is the ball's initial height.)
The height of the goal is 3.048 m. (This is the ball's final height.)


Homework Equations



Vf = Vyo + at
Vyo = VoSinΘ



The Attempt at a Solution



I solved the first part as follows:
Vyo = VoSinΘ
Vf = Vyo + at
0 = 13.5768 - 9.8t
-13.5768 = -9.8t
t = 1.3854s (Time to get to maximum height. This answer is correct.)

I'm having some trouble with the second and third part though. For the second part:
Vyo = VoSinΘ
d = Vyot + 1/2at^2

I subtracted the two heights (3.048m - 2.576m = 0.472m).
0.472m = 13.5768t + 1/2(9.8m/s^2)t^2

I then used the quadratic formula:
-13.5768 +- (square root of: ((13.5768^2) - 4(4.9)(-0.472))/(2)/(-0.472)

The answer I derived was not correct (28.4).

For the third part:

Xmax = 2Vo^2sinΘcosΘ/g

2(17^2)(sin53)(cos53)/9.8

The answer I derived was not correct (28.3).

Please help me if you can!
 
  • #3


I would suggest double checking your calculations and equations to ensure they are correct. It's also important to make sure you are using the correct units and converting them if necessary. It may also be helpful to draw a diagram and label all the given information to visualize the problem better.

For the second part, you can use the equation d = Vyo*t + 1/2at^2 to solve for the time it takes the ball to reach the hoop. You already know the initial velocity (Vyo), acceleration (a), and final height (d). You can plug those values in and solve for t.

For the third part, you can use the equation x = Vxo*t to solve for the horizontal length of the shot. Since the ball is being thrown at an angle, you will need to find the initial horizontal velocity (Vxo) using the equation Vxo = VoCosΘ. You already know the initial velocity (Vo) and the angle (Θ). You can then plug those values in and solve for x.

Remember to always double check your units and use the correct equations for the given scenario. It's also helpful to show your work and label your final answer with the correct units.
 

1. What is a projectile?

A projectile is any object that is launched into the air and moves in a curved path due to the force of gravity.

2. How is the range of a projectile calculated?

The range of a projectile is calculated using the formula R = (v^2 * sin2θ) / g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

3. What is the maximum height of a projectile?

The maximum height of a projectile is the highest point it reaches during its flight. It can be calculated using the formula H = (v^2 * sin^2θ) / (2g), where v is the initial velocity and θ is the launch angle.

4. How is the time of flight for a projectile determined?

The time of flight for a projectile is determined by dividing the range by the horizontal component of the initial velocity, which is v * cosθ.

5. What factors affect the motion of a projectile?

The motion of a projectile is affected by factors such as the initial velocity, launch angle, air resistance, and the acceleration due to gravity. Other factors may include the shape and size of the object and external forces acting on it.

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