Question

You launch a ball at a target with a speed v0 = 10.84 m/s at an angle θ above the horizontal. The target is at a height h = 5.22 m above the level at which you release the ball. You want the ball’s velocity to be horizontal (i.e. vy= 0 m/s) at the instant it reaches the target.

What is the value of the horizontal velocity v0x in m/s?

Enter your value with two (2) decimal places and no units. For example if your answer is 1.276, then enter 1.28.

Answer 1

Step-by-step explanation:

We can use the kinematic equation for vertical motion to find the value of the initial vertical velocity, v0y:

v0y = v0 * sin(θ)

Next, we can use the equation for vertical motion with constant acceleration (g = 9.8 m/s^2 is the acceleration due to gravity) to find the time it takes for the ball to reach the target:

v0y * t - 0.5 * g * t^2 = h

Solving for t:

t = sqrt(2 * h / g)

Finally, we can use the kinematic equation for horizontal motion (no acceleration in the horizontal direction) to find the horizontal velocity:

v0x = v0 * cos(θ)

So,

v0x = 10.84 * cos(θ) = 10.84 * cos(θ)

v0x = 10.84 * cos(θ) = 10.84 * cos(θ)

We do not have the value of θ, so we cannot determine the exact value of v0x.