Question dealing with equations of motion

[bob 911]

There was a recent exam question and i was wondering how to find out the answer.

Q. A basketball player throws a ball with an initial velocity of 6.5m/s at an angle of 50 degrees to the horizontal. The ball is 2.3m above the ground when released. The ball travels a horizontal distance of 2.9m to reach the top of the basket.(see attatchment).

a) Calculate
i) the horizontal component of the initial velocity of the ball
ii) the vertical component of the initial velocity of the ball
b) show that the time taken for the ball to reach the basket is 0.69s
c) Calculate the height H of the top of the basket.

TIA

 

[LowlyPion]

There was a recent exam question and i was wondering how to find out the answer.

Q. A basketball player throws a ball with an initial velocity of 6.5m/s at an angle of 50 degrees to the horizontal. The ball is 2.3m above the ground when released. The ball travels a horizontal distance of 2.9m to reach the top of the basket.(see attatchment).

a) Calculate
i) the horizontal component of the initial velocity of the ball
ii) the vertical component of the initial velocity of the ball
b) show that the time taken for the ball to reach the basket is 0.69s
c) Calculate the height H of the top of the basket.

TIA

Welcome to PF.

OK so where would you think to begin?

For a) i) and ii) you don't have any problem do you?

 

[bob 911]

Thanks for the reply.

A)i) I used f=f1cos(angle)
f= 6.5*cos50
= 4.18m/s

ii) I used f=f1sin(angle)
f= 6.5*sin50
=4.98m/s

B) I used Sh (horizontal distance travelled) = Uh (Horizontal component of the intial velocity) * time
2.9m = 4.18 * T
T = 0.69s

C) This is the part I am not sure about.

My method was to find the height of the basketball at its maximum point of the projectile. (remembering to add my answer to 2.3m since it left from that height)
I then found how much it fell by after the maximum point to the basketball net (taking that away from the maximum height i worked out)
My final answer was 3.0m.

However, i am not sure if this was the way to tackle this problem.

 

[HallsofIvy]

You could use "Sh (horizontal distance travelled) = Uh (Horizontal component of the intial velocity) * time" because there no horizontal component of force and so the horizontal velocity is constant.

There is a vertical force of -mg and so a constant vertical acceleration of -g. You should be able to find formulas for both vertical velocity and height at time t, either by integrating or using given formulas. You know that the time the ball reaches the .69 s. Just put that t into the height formula.