1) A golfer hits a shot to a green that is elevated 3 m above the point where th

Question

1) A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck. The ball leaves the club at a speed of 16.3 m/s at an angle of 43.6° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands. v = 2) On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has x and y components of v0x = 3065 m/s and v0y = 5525 m/s. While the engines are firing, the craft undergoes a displacement that has components of x = 4.84 106 m and y = 6.00 106 m. Find the x and y components of the craft's acceleration. ax = m/s2 ay = m/s2 3) A major-league pitcher can throw a baseball in excess of 42.3 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 15.5 m away from the point of release?

Explanation / Answer

Vx = v0cos = 16.3cos(43.6) = 11.80 m/s
Vy = v0sin = 16.3sin(43.6) =11.24 m/s

y = y0 + v0t + (1/2)at^2
y = 11.24t - 4.9t^2
3 = 11.24t - 4.9t^2
4.9t^2 - 11.24t + 3 = 0
---- Quadratic Formula ---
t = 0.3083 s
t = 1.986 s <=== This is the time we want

Vx = 11.24 m/s (constant)

Vy = 11.24 - 9.8t
Vy = 13.585 - (9.8)(1.986)
Vy = -5.8778 m/s

V^2 = (Vx)^2 + (Vy)^2
V = sqrt((11.24)^2 + (-5.8778)^2)
V = 9.580 m/s


Edit: It would probably be easier to do this as an energy problem:

K = K + P
(1/2)mv0^2 = (1/2)mv^2 + mgh
v0^2 = v^2 + 2gh
v^2 = v0^2 - 2gh
v = sqrt(v0^2 - 2gh)
v = sqrt(16.3^2 - 2(9.8)(3))
v = 14.384 m/s

2.Technically in real life a rocket's acceleration is not constant, but we're going to ignore that little fact for now.

The equation you should be using is x = (1/2)a[x]*t^2 + v[0x]*t
This applies to the y-direction with y = (1/2)a[y]*t^2 + v[0y]*t

x is x-displacement and y is y-displacement. a[x] is x-acceleration and a[y] is y-acceleration. v[0x] is x initial velocity and v[0y] is y initial velocity. Now just solve for a[x] and a[y]:

a[x] = 2(x / t^2) - 2(v[0x] / t)
a[y] = 2(y / t^2) - 2(v[0y] / t)

3. Vox = 42.3 m/s
Voy = 0.0 m/s

Using:
X = Xo + Vox t + (1/2)at^2 <--- General formula

Substituting:
15.5= 0 + (41.0)t + 0
t = 0.378 seconds

Now using:
Y = Yo + Voy t + (1/2)gt^2

Substituting:
Y= 0 + 0 + (4.9)(0.378)^2
y = 0.700 m

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