Question:
A 47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 24.8 m.?
Stephanie
2010-02-20 09:13:06 UTC
What is its speed when it is 6.0 m below its highest point?
Three answers:
Algol
2010-02-21 06:28:49 UTC
We only need to consider the point 6.00m below the high point:



24.8m - 6.00m = 18.8m



Then equate the kinetic energy at the tee with the kinetic plus potential energy at 18.8m above the ground because gravity is a conservative force. From the law of conservation of mechanical energy:



0.5mv²(i) = 0.5mv²(f) + mgh

v(f) = √[v²(i) - 2gh]

= √[(52.0m/s)² -2(9.80m/s²)(18.8m)]

= 48.3m/s



Note that mass divides out so is not needed to find the answer.



Hope this helps.
Mike T
2010-02-20 22:45:26 UTC
The 2 formulas we need:

Position as a function of time:

p(t) = 0.5 * a * t^2 + v0 * t + p0

Velocity as a function of time:

v(t) = a * t + v0



We are given the initial speed off the tee, but not the angle, so we don't know the horizontal and vertical components (vx0, vy0).



Use -9.8 m/sec^2 as the acceleration due to gravity.



When the ball is at the apogee the vertical component of the velocity will be 0, so

v(t) = 0 = -9.8 * t + vy0

9.8 * t = vy0

t = vy0 / 9.8



At that same time

P(t) = 24.8 = 0.5 * -9.8 * t^2 + vy0 * t + 0

24.8 = -4.9 * t^2 + vy0 * t

Substituting

24.8 = -4.9 * (vy0/9.8)^2 + vy0 * vy0/9.8

24.8 = -4.9 * (vy0/9.8)^2 + vy0 * vy0/9.8

24.8 = -vy0^2 / 19.6 + vy0^2 / 9.8

486.08 = -vy0^2 + 2 * vy0^2

486.08 = vy0^2

22.047222 = vy0



Because the initial speed of the ball is 52.0,

(vx0, vy0) = (52.0 * cos(p), 52.0 * sin(p))

22.047222 = vy0 = 52.0 * sin(p)

0.423985 = sin(p)

arcsin(0.423985) = p

25.0864 = p

vx0 = 52 * cos(25.0864)

vx0 = 52 * 0.905669

vx0 = 47.095



Now that we know the initial velocity (speed plus direction), the next question is how long does it take to get to the height 6 meters below the apogee?



py(t) = (24.8 - 6) = 0.5 * -9.8 * t^2 + 22.047222 * t + 0

18.8 = -4.9 * t^2 + 22.047222 * t

0 = -4.9 * t^2 + 22.047222 * t - 18.8



Use the general quadratic equation and you get two values for t, 1.14314987 or 3.356283191



We use the first, smaller value, because that represent the time it take to reach the desired height as it travels upward to the apogee.



To find the vertical component of the velocity at that time

vy(1.239563997) = -9.8 * 1.14314987 + 22.047222 = 10.844353274



The horizontal component of the velocity will be vx0 because there is no acceleration in the horizontal direction. Therefore the velocity is (47.095, 10.844). The speed is the magnitude of the vector



s = sqrt(47.095^2 +10.844^2)

s = 48.3 m/s
Physicsquest
2010-02-21 05:45:26 UTC
v^2 = 2gh



2gh = 2 x 9.8 x 24.8 = 486.08, sq-root = initial vertical velocity = 22.05 m/s



arcsin(22.05/52) = 25.09 degrees from horizontal



horizontal velocity = 52cos(25.09) = 47.093 m/s



v^2 = 2gh

v^2 = 2 x 9.8 x 6 = 117.6, sq-root = v = 10.84 m/s



10.84^2 + 47.093^2 = answer^2 = 2335.256, sq-root = v = 48.32 m/s


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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