A football is kicked at 37degrees to the horizontal and it goes a distance of 98 yd. Ignore air resistance. a) what is the ball’s initial velocity? b) what is the vertical component of the ball’s velocity? c) what is the horizontal component of the ball’s velocity d) how long is the ball in the air? (Choose positive vertical direction upwards)
If the same football was punted with the same initial velocity (from question a) on the surface of the moon where a=1.62m/s^2 how long does it take the ball to return to the ground? What is the max height? What is the horizontal range?
98yd=89.61m
a) delta x=v_0^2/g
sqrt((89.61)(10))=29.9m/s
b) voy=v_0sin theda
=(29.9)sin37
=17.99
c) vox=v_0cos theda
=(29.9)cos37
=23.87m/s
d) t=d/vx
=89.61/23.87
=3.75s
For where the moon has a=1.62m/s^2, wouldn’t time also equal 3.75s
for the max height:
d =voy(t) +1/2at^2
=17.99*1.875=1/2 (-1.62)(1.875)^2
=33.73+ -2.84
=30.89m
(would it be -1.62 or would it be positive if we choose vertical direction to be up?)
the range would be:
delta x=vx(t)
=(23.87)(3.75)
=89.51m
Are these correct?
A football is kicked at 37 degrees to the horizontal and it goes a distance of 98 yd. Ignore air resistance. a) what is the ball’s initial velocity? b) what is the vertical component of the ball’s initial velocity? c) what is the horizontal component of the ball’s initial velocity d) how long is the ball in the air? (Choose positive vertical direction upwards)
a--The horizontal distance traveled is 98(3) = 294 feet.
d = V^2(sin2µ)/g where d is the horizontal distance traveled in feet, V is the initial velocity in ft./sec., µ is the velocity angle with the horizontal and g = the acceleration due to gravity, 32.2 ft./sec.^2.
Therefore, using d = V^2sin(2µ)/g, 294 = V^2(sin74)/32.2 = from which Vo = 99.24 ft/sec.
b--The vertical component of the initial velocity is 99.24(sin37) = 59.72 ft/sec. ft/sec.
c--The horizontal component of the initial velocity is 99.24(cos37) = 79.25 ft/sec.
d--From Vf = Vo - gt where Vo = the initial vertical velocity component, Vf = the final vertical velocity component (= 0) and t = the time from time = 0 to the maximum height at t(up), 0 = 99.23 - 32.2t(up), t(up) = 99.23/32.2 = 3.08 sec.
Since it takes the same amount of time to fall back to the ground, the total time in the air is 6.16 sec.
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Given the above, you should be able to determine the moon punt answers.
Let's break down each part of the question and go through the calculations to determine if the answers provided are correct.
a) To find the ball's initial velocity, we can use the formula: Δx = v0^2 / g.
Given that the distance traveled is 98 yards or 89.61 meters, we can rearrange the formula and solve for v0:
v0 = sqrt(Δx * g) = sqrt(89.61 * 10) = 29.9 m/s
Therefore, the initial velocity of the ball is 29.9 m/s.
b) To find the vertical component of the ball's velocity, we can use the formula: voy = v0 * sin(θ).
Using the value of v0 calculated in part a) and the given angle of 37 degrees, we can evaluate the equation:
voy = 29.9 * sin(37) = 17.99 m/s
So, the vertical component of the ball's velocity is 17.99 m/s.
c) To determine the horizontal component of the ball's velocity, we use the formula: vox = v0 * cos(θ).
Again, using the value of v0 from part a) and the angle of 37 degrees, we can calculate:
vox = 29.9 * cos(37) = 23.87 m/s
The horizontal component of the ball's velocity is 23.87 m/s.
d) To find how long the ball is in the air, we can use the formula: t = Δx / vx.
Given that the distance traveled is 89.61 meters (from part a) and the horizontal component of the velocity is 23.87 m/s, we can calculate:
t = 89.61 / 23.87 = 3.75 seconds
So, the ball is in the air for 3.75 seconds.
Now, moving on to the second part of the question, where we consider the moon's gravity (a = 1.62 m/s^2).
For the time it takes for the ball to return to the ground, it will still be 3.75 seconds. The acceleration due to gravity does not affect the time of flight.
To calculate the maximum height, we can use the equation: d = voy * t + (1/2) * a * t^2.
Plugging in the values of voy (17.99 m/s), t (3.75 s), a (-1.62 m/s^2), and solving the equation, we obtain:
d = 17.99 * 3.75 + (1/2) * (-1.62) * (3.75)^2 = 30.89 m
The maximum height reached by the ball is 30.89 meters.
The horizontal range remains the same as in part d), which is 89.51 meters.
Therefore, based on the calculations provided, the answers for each part of the question are correct.