Projectile Motion - find initial velocity, flight time and max height

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a golf ball shot from a height of 6.5 ft at a 45° angle. The initial velocity was calculated to be 10.82 m/s using the range formula R = (v0cosθ)*[(2v0sinθ)/g]. The flight time was determined to be approximately 23.9 seconds, with additional calculations needed to find the time for the ball to drop the remaining height. The maximum height reached by the ball was computed as 4.96 m.

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  • Familiarity with unit conversions (imperial to metric)
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kgal
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Homework Statement


a golf ball is shot from a height of 6.5 ft above the ground at an angle of 45° above the horizontal. The ball lands 39 ft and 3 inches downfield. Assuming ideal projectile motion find:
a. Initial velocity
b. flight time
c. maximum height


Homework Equations


r = [(v0*cosθ)t ]i + ((v0*sinθ)t - 1/5gt^2)j.


The Attempt at a Solution


I have attached my work (problem 5) but can't seem to know if it correct. I have converted all of my units to metric because gravity is know as 9.8 m/s^2.
 
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kgal,
From what I can gather, you don't have enough information given to you to solve for velocity, time of flight and maximum height directly. Try drawing what you see. The best method I use to solve projectile motion problems is to split the problem up into two parts - the object rising, & then falling.

Also, I can't see your work.
 
I used the definition of range to solve for initial velocity:
a.R = (v0cosθ)*[(2v0sinθ)/g]
solved for v0 = sqrt (Rg / 2cosθsinθ) = 10.82 m/s.

b. I split the problem into two pieces, the time it takes the ball to reach the horizontal
t1 = 2v0sinθ / g = 23.9 s
Then got stuck on finding out the time it took the ball to drop the remaining 1.98 m...
I tried finding t2 by using r = (v0cos45)t + (v0sin45t - 1/2 gt^2) and solve for t, but got to the point where t was a quadratic equation with 2 answers, t = 3.1 or 0.13 seconds...

c. y max = y0 + (v0sinθ) ^2 / 2g = 4.96 m
 

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