Projectile motion - finding initial velocity with no time

[KKuff]

Homework Statement

I'm having trouble finding the initial velocity without knowing the time.

A basketball player is standing on the floor 10.0 m from the basket. The height of the basket is H = 3.05 m, and he shoots the ball at an angle, θ = 42.0°, with the horizontal from a height of h = 1.94 m. At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?

Homework Equations

I'm not sure how to determine which equations I need, since there are at least 2 variables that I don't know in each one

V(f) = V(i) + at
X(f) = X(i) + (1/2)(V(i) + V(f))t
X(f) = X(i) + V(i)t + (1/2)at^2
V(f)^2 = V(i)^2 + 2a(X(f) - X(i))

The Attempt at a Solution

I know that I have to break up the problem into x and y components. So I would have the y-velocity component Vy = Vsin42.0 and the x-velocity component Vx = Vcos42.0. I know that for the y component I would have an acceleration of -9.8m/s and that for the x component that I would have no acceleration since it is a constant velocity, so that would make Vx(i) = Vx(f). The initial horizontal position is X(i) = 0 and the final horizontal position is X(f) = 10. The initial vertical position is Y(i) = 1.94 and the final vertical position is Y(f) = 3.05.

I just don't know how to put all of this information together in order to find the initial velocity. Any help will be appreciated.

 

[tiny-tim]

Welcome to PF!

Hi KKuff! Welcome to PF!

(try using the X² and X₂ icons just above the Reply box )

Call the speed v, then find t from the x equation, and use that value of t in the y equation, to solve for v …

what do you get? ​

 

[KKuff]

I got about 10.6m/s. Thanks I appreciate it. Is t used since it would have the same value in both the x and the y components?

 

[tiny-tim]

Is t used since it would have the same value in both the x and the y components?

That's right!

(standard technique … sometimes it's t, sometimes it's x or y or even v )

 

[rwisz]

To find the initial velocity in this problem, we can use the kinematic equations for projectile motion. However, as you mentioned, there are two unknown variables (time and initial velocity) in each equation, so we will need to use multiple equations and solve for both variables simultaneously.

First, we can use the equation X(f) = X(i) + Vx(i)t to solve for the time it takes for the ball to travel 10.0 m horizontally. Since the initial horizontal position is 0, this equation becomes X(f) = Vx(i)t. Plugging in the values, we get 10.0 m = Vcos42.0 t. We can rearrange this to solve for t, which gives us t = 10.0 m / (Vcos42.0).

Next, we can use the equation Y(f) = Y(i) + Vyi t + (1/2)at^2 to solve for the initial vertical velocity, Vyi. Since we want the ball to go through the hoop without hitting the backboard, the final vertical position should be equal to the height of the hoop, 3.05 m. Plugging in the values, we get 3.05 m = 1.94 m + Vsin42.0 t - (1/2)(9.8 m/s^2)(10.0 m / (Vcos42.0))^2. We can rearrange this to solve for Vyi.

Finally, we can use the Pythagorean theorem to find the magnitude of the initial velocity. V = √(Vxi^2 + Vyi^2). Plugging in the values for Vxi and Vyi that we have solved for, we can find the initial velocity needed to make the shot.

It is important to note that there may be some trial and error involved in solving this problem, as we are using multiple equations and variables. It is always a good idea to check your final answer by plugging it back into the original equations to make sure it satisfies all of the given conditions.