Projectile motion tennis ball problem

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SUMMARY

The projectile motion problem involves a tennis ball hit with a vertical speed of 10 m/s and a horizontal speed of 30 m/s. The ball remains in the air for 2 seconds and travels a horizontal distance of 60 meters. To calculate the time in the air, the equation d = v1*t + (1/2)at^2 is applied for vertical displacement, considering the vertical acceleration due to gravity as -9.8 m/s². The horizontal displacement is determined using the constant horizontal velocity over the calculated time.

PREREQUISITES
  • Understanding of basic physics concepts related to projectile motion
  • Familiarity with kinematic equations, specifically d = v1*t + (1/2)at^2
  • Knowledge of vertical and horizontal velocity components
  • Basic grasp of gravitational acceleration (-9.8 m/s²)
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to analyze projectile motion at different launch angles
  • Explore the effects of air resistance on projectile trajectories
  • Investigate real-world applications of projectile motion in sports and engineering
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in understanding the dynamics of sports physics, particularly in tennis.

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Homework Statement


Tennis Ball hit at a vertical speed of 10 m/s & Horizontal speed of 30 m/s. How long will the ball remain in the air? How far will the ball travel horizontally?


Homework Equations


I am not sure which equation to use? The angle that the ball was hit is not given, would I try all angles? 30,37,45,57,60? I know that 45 would give the max distance


The Attempt at a Solution


Answers given:
How far: 60m
How long: 2s

Thanks!
 
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To find how long the ball is in the air, use the equation:

d = v1*t + (1/2)at^2. apply this equation for vertical displacement...
 
As vertical and horizontal components of velocities are given, angle is not required. At the highest point the velocity will be zero (v = 0). Initial vertical velocity = u = 10 m/s. Acceleration a = - 9.8 m/s^2 (-ve because acceleration due to gravity is always downwards i.e. opposite of the +ve upwards displacement) Apply v = u + at to get t, the time to reach the highest point. Double that to get the time during which the ball is in air (T).

As the horizontal velocity remains constant throughout, determine horizontal displacement in time T (calculated above) using s = ut
 

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