Projectile Motion: Angle & Distance of a Baseball Leaving a Bat at 50 m/s

AI Thread Summary
A baseball leaves a bat at a speed of 50 m/s and reaches its highest point in 4.42 seconds. To determine the angle of projection, the vertical component of velocity at the peak must be zero, allowing for calculations using projectile motion equations. The angle can be found using the relationship between the initial vertical and horizontal velocities. Once the angle is established, the horizontal component of the velocity can be used to calculate the shortest distance from the starting point to the highest point of the ball's trajectory. Understanding these concepts is essential for solving basic projectile motion problems.
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A baseball leaves a bat at a rate of 50 m/s. The ball reaches its highest point after 4.42 seconds.
(a) What is the angle with the horizontal at which the ball leaves the bat?
(b) What is the shortest distance between the starting point and the highest point of the ball?

Attempt:
Vfy = Viy + a \times \Deltat
:confused:
 
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This is a fairly basic projectile motion question, so you might want to brush up the concepts again.
Yes, use vy=uy+ayt
So the question here is what is vy at the topmost point. After you find uy, you can easily find the angle of projection using concepts of vectors.
The second part follows easily once you have the horizontal component of the velocity.
 
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