A basketball is thrown at 45° to the horizontal. The hoop is located 6.2 m away horizontally at a height of 1.1 m above the point of release. What is the required initial speed?

I know that I need to use d = vit + (0.5)at^2, but I then get stuck. Thanks!

Y^2 = Yo^2 + 2g*h = 0

Yo^2 - 19.6*1.1 = 0
Yo^2 = 21.56
Yo = 4.64 m/s.

Vo = Yo/Sin A = 4.64/sin45 = 6.57 m/s @
45o.

Well, well, well! Looks like we've got a basketball enthusiast here. Let me try to help you calculate that initial speed, my friend.

First things first, we can break down the motion of the basketball into its horizontal and vertical components. Since we're looking for the initial speed, we'll only focus on the vertical component.

Now, considering that the hoop is at a height of 1.1 m above the release point, we can use the vertical displacement (d) to find the time it takes for the basketball to reach the hoop. In this case, our displacement is 1.1 m, and we don't have any initial or final velocities in the vertical direction, so we can use the equation:

d = vit + (0.5)at²

Since the basketball is thrown horizontally at 45°, the initial vertical velocity (viy) is 0 m/s. Yup, that's right, zero. Gravity, my friend, doesn't impact horizontal movement.

So, our equation becomes:

1.1 = 0 * t + (0.5) * (-9.8) * t²

Where -9.8 m/s² represents the acceleration due to gravity. Simplify that equation, solve for t, and you'll get the time it takes for the basketball to reach that height of 1.1 m.

Once you have the time, you can use the horizontal displacement of 6.2 m and the known time to calculate the initial horizontal velocity (vix). Remember, the horizontal velocity remains constant throughout the motion, so:

vix = dx / t

Plug in the values, solve for vix, and you've got your horizontal velocity.

Now, to find the initial speed, we need to combine the horizontal and vertical velocities using the Pythagorean theorem:

vi = √(vix² + viy²)

And voilà! You've found your required initial speed to make that epic hoop shot. I hope this explanation made it more bearable than watching a clown play basketball!

To determine the required initial speed of the basketball, you can break down the problem into horizontal and vertical components.

First, let's consider the horizontal motion. The distance traveled horizontally is given as 6.2 m. The horizontal component of velocity remains constant throughout the motion, as there is no horizontal acceleration. Therefore, we can use the equation:

d = v_i * t

where d is the horizontal distance traveled, v_i is the initial horizontal velocity, and t is the time of flight.

In this case, since the basketball is thrown at 45° to the horizontal, the initial horizontal velocity (v_i) can be calculated as:

v_i = v * cos(theta)

where v is the initial speed (which we want to find) and theta is the launch angle. In this case, theta = 45°.

Now, let's consider the vertical motion. The basketball is thrown upward and then falls back down, so we can use the vertical displacement formula:

h = v_i * t + (0.5) * a * t^2

where h is the change in height, v_i is the initial vertical velocity, t is the time of flight, and a is the vertical acceleration due to gravity (-9.8 m/s^2).

In this case, the vertical displacement (h) is given as 1.1 m above the release point, and the initial vertical velocity can be calculated as:

v_i = v * sin(theta)

Now, we can set up a system of equations:

Equation 1: d = v_i * t
Equation 2: h = v_i * t + (0.5) * a * t^2

Substituting the expressions for v_i from the earlier equations:

Equation 1: d = v * cos(theta) * t
Equation 2: h = v * sin(theta) * t + (0.5) * a * t^2

With the given values of d = 6.2 m and h = 1.1 m, and theta = 45°, we have a system of two equations with two unknowns (v and t).

Now, solve Equation 1 for t:

t = d / (v * cos(theta))

Substitute this value into Equation 2:

h = (v * sin(theta) * d) / (v * cos(theta)) + (0.5) * a * (d^2 / (v^2 * cos^2(theta)))

Now you have an equation with one unknown, v. Simplify and solve for v:

h = sin(theta) * d / cos(theta) + (0.5) * a * (d^2 / (v * cos^2(theta)))

v * cos(theta) = d / t [from Equation 1]

h = sin(theta) * d / cos(theta) + (0.5) * a * (d^2 / ((d / (v * cos(theta))) * cos^2(theta)))

simplify further...

Now, you can plug in the known values of d, h, theta, and a to solve for v. Rearrange the equation and solve for v:

v = sqrt((d * g * sin(2*theta)) / (2 * (cos(theta))^2 * (d * tan(theta) - h)))

By substituting the given values of d = 6.2 m, h = 1.1 m, theta = 45°, and g = 9.8 m/s^2, you can calculate the required initial speed (v) using the formula above.