A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How long was the ball in the air? Round the answer to the nearest tenth of a second.
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To find out how long the ball was in the air, we can use the vertical motion equation:
h = v_initial * t - (1/2) * g * t^2
where:
h = height (which is 0 for the entire trajectory since the ball lands at the same height it was hit)
v_initial = initial velocity (31 m/s)
t = time
g = acceleration due to gravity (9.8 m/s^2)
Since the ball lands at the same height it was hit, we can set h = 0 and solve for t:
0 = v_initial * t - (1/2) * g * t^2
Rearranging the equation, we have:
(1/2) * g * t^2 = v_initial * t
Multiplying both sides by 2, we get:
g * t^2 = 2 * v_initial * t
Dividing both sides by t and rearranging the equation, we have:
g * t = 2 * v_initial
Now, we can substitute the values:
9.8 * t = 2 * 31
Dividing both sides by 9.8, we get:
t = (2 * 31) / 9.8
Calculating the expression, we find:
t ≈ 6.33 seconds
Rounding the answer to the nearest tenth of a second, the ball was in the air for approximately 6.3 seconds.
To find the time the ball was in the air, we can use the horizontal component of its velocity. Let's break down the initial velocity into its horizontal and vertical components.
Given:
Initial speed (velocity) of the ball, v = 31 m/s
Angle of projection, θ = 35 degrees
The horizontal component of velocity (Vx) can be calculated using the formula:
Vx = v * cos(θ)
Let's plug in the values and calculate Vx:
Vx = 31 * cos(35)
Using a scientific calculator, we find that Vx ≈ 25.4 m/s (rounded to one decimal place).
Next, let's calculate the time the ball is in the air.
We can use the formula for vertical displacement:
Δy = v * t * sin(θ) - 1/2 * g * t^2
In this case, the vertical displacement (Δy) is 0 because the ball lands at the same vertical level as it started.
Therefore, the formula simplifies to:
0 = v * t * sin(θ) - 1/2 * g * t^2
The value of g (acceleration due to gravity) is approximately 9.8 m/s^2.
We can rearrange the equation to solve for time (t):
1/2 * g * t^2 = v * t * sin(θ)
Simplifying further:
1/2 * 9.8 * t^2 = 31 * t * sin(35)
Dividing both sides by 15.5 (to simplify the equation further):
t^2 = (31 * sin(35)) / 15.5
Now, let's solve for t by taking the square root of both sides:
t = √[(31 * sin(35)) / 15.5]
Using a scientific calculator, we find that t ≈ 2.2 seconds (rounded to one decimal place).
Therefore, the ball was in the air for approximately 2.2 seconds.
In projectile motion,
h,max = vo^2 sin^2 (x) / 2g
where
h,max = maximum height
vo = initial velocity
x = angle of release
g = acceleration due to gravity = 9.8 m/s^2
Solving for the maximum height,
h,max = 31^2 * sin^2 (35) / 2(9.8)
h,max = 316.16 / 19.6
h,max = 16.13 m
We can now solve for t using the equation:
h = vo,y*t - (1/2)gt^2
Substituting,
16.13 = 31*sin(35) t - 4.9 t^2
4.9t^2 - 17.78t + 16.13 = 0
This is obviously a quadratic equation, and so we use quadratic formula. getting the roots, we'll have
t = 1.81 s
This time is the time the object reaches the maximum height. Since the problem asks for the time the object is in air, we multiply is by 2 to account for the time it falls back the ground.
t = 3.62 s
Actually, there's really a formula to solve for time of flight, which is
t = 2 vo sin(x) / g
If you use this, you should get approximately the same answer. What I did above is the longer method.
hope this helps~ `u`