What is the initial projectile speed of the ball?

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SUMMARY

The initial projectile speed of the ball in the slow-pitch softball game can be calculated using kinematic equations. The ball is released from a height of 3.0 ft, and the time to reach maximum height is half of the total flight time. The correct approach involves using the equation for vertical motion, taking into account the initial height and the acceleration due to gravity (32 ft/s²). The maximum height is determined by adding the initial height to the calculated height gained during the ascent.

PREREQUISITES
  • Kinematic equations of motion
  • Understanding of projectile motion
  • Basic algebra for solving equations
  • Knowledge of gravitational acceleration (32 ft/s²)
NEXT STEPS
  • Calculate the initial speed of the ball using the formula: v = u + at
  • Determine the time to reach maximum height using t = v/g
  • Learn how to apply the kinematic equation for vertical displacement: Δy = ut + 0.5at²
  • Explore the concept of maximum height in projectile motion and its calculation
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Students studying physics, coaches analyzing softball pitching techniques, and anyone interested in understanding projectile motion dynamics.

kappcity06
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The pitcher in a slow-pitch softball game releases the ball at a point 3.0 ft above ground level. A stroboscopic plot of the position of the ball is shown in the figure below, where the readings are 0.25 s apart and the ball is released at t = 0. On the horizontal axis, x1 = 20 ft.


(a) What is the initial speed of the ball?
ft/s
(b) What is the speed of the ball at the instant it reaches its maximum height above ground level?
ft/s
(c) What is that maximum height?
ft


i tried to find part c first. since i know that time is equal to 1.25 seconds i used deltay=.5at^2
.5(32)(1.25)^2
25 feet
this did not work so I do not know what to do?
any one with any help. thaks
 

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Why are you using t=1.25s in your calculation? That isn't the time it takes the ball to get to maximum height.
 
If you can answer (b), that will help you answer (c).

Dorothy
 
u=0 i think
 
Gib Z said:
u=0 i think

u can't be zero as the ball is pitched!
 
i am still not getting the answer. I also tried time as 1.25/2 but this is stil not right?

any one please
 
1.25/2 is the time to the maximum height. The equation you were using gives the change in height from where it started. Don't forget it was 3 ft above the ground when it was thrown. So if you want the total height above the ground you have to take that into account.

Does that help?
 

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