Calculating Distance and Height of Ball w/ 18° Loft & 10.9s Air Time

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SUMMARY

This discussion focuses on calculating the distance and maximum height of a golf ball hit with an 18° loft and an initial velocity of 18.5 m/s, remaining airborne for 10.9 seconds. The horizontal distance is determined using the equation deltaX = Vx(deltaT), where Vx is the horizontal component of the initial velocity. The maximum height is calculated using kinematic equations, considering the vertical acceleration of 9.8 m/s² and the initial vertical velocity derived from the loft angle.

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Homework Statement


If a wood with loft 18 degrees is used to hit a ball that is in the air for 10.9 s, calculate:
(a) The distance the ball travels if its intial velociity is 18.5 m/s.
(b) The maximum height of the ball


Homework Equations





The Attempt at a Solution

 
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@d@m said:

Homework Statement


If a wood with loft 18 degrees is used to hit a ball that is in the air for 10.9 s, calculate:
(a) The distance the ball travels if its intial velociity is 18.5 m/s.
(b) The maximum height of the ball


Homework Equations





The Attempt at a Solution


For a:
If you know the time in which the ball is in the air for you can easily find the horizontal range. You do this because you know the initial velocity, which is 18.5 m/s [18 deg above the horizontal]. You can calculate the horizontal speed, which is always constant in this case. Then use the equation deltaX=Vx(deltaT) to find the horizontal range...

For b:
Time is independent of the vertical and horizontal directions so you can use the same time in the vertical calculations... You know acceleration in the y direction (9.8), You know the initial speed in the y direction because you can easily calculate it... And if you want to know the maximum height of the ball you know that it's speed at the top of its flight will be 0... Now you have 3 pieces of info you can use to solve b...

Hope I helped...
 

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