Baseball Velocity Projectile motion

AI Thread Summary
To determine how far a baseball thrown horizontally at 100.8 mi/hr travels vertically by the time it reaches home plate 60.0 ft away, the time taken must be calculated using the formula t = d/v. For the cliff scenario, a rock thrown horizontally at 18.0 m/s from a height of 50 m requires finding the time to fall using the vertical displacement formula d = Vi*t + 1/2*a*t^2, where the initial vertical velocity is zero. Both problems involve understanding horizontal and vertical motion separately, with gravity affecting vertical displacement. The final velocities and angles of impact can be calculated once the time of flight is determined. Both scenarios highlight the importance of time in projectile motion calculations.
cowgiljl
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I just need alittle help on getting started with just the formulas.
There are just two question.

1) A baseball thrown at 100.8 mi/hr. If the pitch were thrown horizonally with this velocity how far would the ball travel vertically by the time it reaches home plate 60.0 ft away?

Do i need to worry about the time it takes?

2) If a student stands at the edge of a cliff and throws a rock horizonally over the edge with a speed of 18.0 m/s, The cliff is 50 m above a flat. How long after being released does the rock strile the beach below the cliff? With what speed and angle of the impact does it land.

I do know the the angle = tan-1 (Vy/Vx)

I am just struggling a bit on these two.

Thanks
 
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1) A baseball thrown at 100.8 mi/hr. If the pitch were thrown horizonally with this velocity how far would the ball travel vertically by the time it reaches home plate 60.0 ft away?

Do i need to worry about the time it takes?

Well, yes, you do. You know the horizontal velocity and that there is (neglecting air resistance) no horizontal acceleration so it should be easy to determine the time a ball moving at 100.8 mi/hr takes to travel 60.0 feet (you might want to convert that speed to ft/sec). You know that there is no initial vertical speed and the vertical acceleration is -32.2 ft/sec2.

2) If a student stands at the edge of a cliff and throws a rock horizonally over the edge with a speed of 18.0 m/s, The cliff is 50 m above a flat. How long after being released does the rock strile the beach below the cliff? With what speed and angle of the impact does it land.

I do know the the angle = tan-1 (Vy/Vx)

Once again, you know that the horizontal speed, 18.0 m/s and that there is no acceleration vertically. You know that there is no initial vertical speed and the vertical acceleration is -9.8 m/s2. You should be able to write down the horizontal and vertical speed at any time t as well as the horizontal and vertical distance from the initial point. Since the rock has to go down 50 m in order to hit the beach, you need to find the time required for the vertical
distance to be -50- then use that to find the speed and angle.
 


1) To solve this problem, we can use the formula for horizontal displacement, which is given by d = v*t, where d is the distance, v is the velocity, and t is the time. In this case, we know the velocity (100.8 mi/hr) and the distance (60.0 ft). However, we do not have the time. This is where we can use the formula for time, t = d/v, where t is the time, d is the distance, and v is the velocity. Since we are looking for the vertical displacement, we can use the vertical velocity formula, given by Vf = Vi + at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. We know that the initial velocity is 0, since the ball is thrown horizontally. We also know that the final velocity is 0 when the ball reaches the ground. Therefore, we can rearrange the formula to solve for time, t = Vf/a. Once we have the time, we can plug it into the horizontal displacement formula to find the vertical displacement.

2) To solve this problem, we can use the formula for horizontal displacement, which is given by d = v*t, where d is the distance, v is the velocity, and t is the time. In this case, we know the velocity (18.0 m/s) and the distance (50 m). However, we do not have the time. This is where we can use the formula for time, t = d/v, where t is the time, d is the distance, and v is the velocity. Once we have the time, we can use the formula for vertical displacement, given by d = Vi*t + 1/2*a*t^2, where d is the distance, Vi is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. We know that the initial velocity is 0, since the rock is thrown horizontally. We also know that the final displacement is -50 m, since the rock is 50 m above the ground. Therefore, we can rearrange the formula to solve for time, t = sqrt(2*d/a). Once we have the time, we can use the formula for final velocity, Vf = Vi + at, to find the final velocity. We
 
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