Projectile motion question ( basketball being thrown into a hoop)

In summary, the question asks to calculate the initial speed of a basketball shot that lands in a hoop 3m high when released from a height of 2.1m and 4.3m away at an angle of 55°. Using the equations for initial horizontal and vertical velocity, and the resultant initial velocity, the attempt at a solution involved finding the vertical displacement of the ball, assuming a vertical velocity of 0 at the moment it went through the hoop, and using the equation for displacement to find time. This approach led to an incorrect answer as the assumption of a zero vertical velocity at that moment was incorrect.
  • #1
mihir23
2
0

Homework Statement


A basketball shoots a ball at an angle of 55° which is 4.3m away( horizontal dist). If the ball is released from a height of 2.1m and lands in the hoop, which is 3m high. Calculate the initial speed of the ball for this shot to be successful.


Homework Equations


initial horiz velocity = v cos(θ)
initial vert velocity = v sin(θ)
Resultant initial velocity = Sqroot(initial horiz v ^2 + initial vert v^2)

The Attempt at a Solution


I decided that the vertical displacement the ball travels is 0.9 m up(3-2.1m)
The question does not give the max height , so i was confused about how to approach it

What i did:
a = -9.8
u= ?
v^2 =u^2 +2as
u^2 =17.64
initial vert velocity = v sin(θ)
therefore
v(initial vert) = 4.2/sin 55
4.2 = v sin 55
= 5.127
v(initial horiz)=5.127/tan55
=3.59
total initial v = Sqroot(5.127^2 + 3.6^2)
=6.26 m/s

This answer is wrong according to the back of my book. Could you please tell me how i have done it incorrectly. If you guys really want it, then i'll post the book's answer.
 

Attachments

  • physics question diagram.jpg
    physics question diagram.jpg
    10 KB · Views: 1,216
Physics news on Phys.org
  • #2
0.9m is not maximum height, so u^2 is not 17.64, but you don't need to find max height anyways.

Start with the displacement equations for both components, then put those equations together to find time.
 
  • #3
Look's like you assumed that the vertical component of velocity is zero at the instant the ball goes through the hoop. You can't impose that condition.
 

FAQ: Projectile motion question ( basketball being thrown into a hoop)

1. How does the angle of release affect the trajectory of a basketball?

The angle of release directly affects the trajectory of a basketball. The higher the angle of release, the higher the ball will go and the shorter the distance it will travel. Conversely, a lower angle of release will result in a lower arc and a longer distance.

2. What is the optimal angle for shooting a basketball into a hoop?

The optimal angle for shooting a basketball into a hoop is approximately 45 degrees. This angle allows for enough height to clear the rim, while also providing enough forward momentum to reach the hoop.

3. How does the initial velocity of the basketball affect its trajectory?

The initial velocity of the basketball, or the speed at which it is thrown, also plays a significant role in its trajectory. The faster the initial velocity, the farther the basketball will travel before reaching the hoop. However, if the velocity is too high, it may cause the ball to bounce off the backboard or rim.

4. Does air resistance affect the trajectory of a basketball?

Yes, air resistance can affect the trajectory of a basketball. As the ball travels through the air, it experiences an opposing force from air molecules, which can slow it down and alter its path. However, for most basketball shots, the effect of air resistance is minimal.

5. How does the height of the hoop affect the trajectory of a basketball?

The height of the hoop does not significantly affect the trajectory of a basketball. As long as the basketball is thrown with the correct angle and velocity, it will follow a similar path regardless of the height of the hoop. However, a higher hoop may require a slightly higher initial velocity to reach it.

Back
Top