Baseball Velocity Projectile motion

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SUMMARY

This discussion focuses on solving projectile motion problems involving horizontal throws. The first problem involves a baseball thrown at 100.8 mi/hr, requiring the calculation of vertical displacement over a horizontal distance of 60.0 ft, utilizing the formula for time, t = d/v, and vertical acceleration of -32.2 ft/sec². The second problem addresses a rock thrown horizontally at 18.0 m/s from a 50 m cliff, where the time to impact and the final velocity and angle are determined using the equations of motion, including d = Vi*t + 1/2*a*t² and Vf = Vi + at.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with kinematic equations for horizontal and vertical motion.
  • Knowledge of unit conversions between miles per hour and feet per second, as well as meters per second.
  • Ability to apply trigonometric functions, specifically the tangent function for angle calculations.
NEXT STEPS
  • Study the kinematic equations for projectile motion in detail.
  • Learn about unit conversions between different measurement systems (imperial and metric).
  • Explore the effects of air resistance on projectile motion for more complex scenarios.
  • Practice solving additional problems involving horizontal throws and vertical drops.
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Students studying physics, educators teaching projectile motion concepts, and anyone interested in applying mathematical principles to real-world scenarios involving motion.

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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