A man stands on the roof of a 15m tall building and throws a rock with a velocity of magnitude 30m/s at an angle of 33deg above the horizontal. Ignore air resistance. Calculate:
a) the maximum heigh above the roof reached by the rock
b) the magnitude of the velocity of the rock just before it strikes the ground
c) the horizontal range from the base of the building to the point where the rock strikes the ground
My work so far:
a) I found the initial x and y components to be Vx = 25.16 and Vy = 16.33
Then I found the time it took for the rock to reach its heightest point:
Vf = Vi + a*t
0 = (16.33) + (-9.8)(t)
t = 1.6s
then I plugged that in to find the height
H_max = [(16.33) + 0]/2 * (1.6) = 13.55 m
b) here is where I'm a little lost
I set up this equation to find the time it takes for the rock to go through its entire fall from the building's top to the ground:
y = vy*t + 1/2*g*t
-15 = (16.33)*t - 4.9*t^2
Then I used the quadratic equation to find the time is t = 3.98s
I plugged that back into the Vf = Vi + a*t equation to find the final velocity
Vf = Vi + a*t
Vf = (16.33) + (-9.8)(3.98)
Vf = -22.67
magnitude is the absolute value of this = 22.67 m/s
But my solution book says that the answer is 34.6 m/s so what part did I do wrong?
c) I used 3.98s to plug into
x = (25.16)*(3.98)
x = 100.64 m
Am I doing the problem correctly? I'm fairly sure that I am for the most part.
a) h = V^2(sin^2(33))/2(9.8)
b) V = 33m/s (no atmosphere)
c)Determine time "t" to fall from building roof height to ground from 15 = 16.33t + 9.8t^2/2
Horizontal distance traveled from building base to impact on ground is
d = V^2(sin(2µ))/g + 25.16t
you need to make your g positive and keep the initial position negative to get the correct t (.704s)
then that gives you your Voy.(21.9)
your Vox = 25.16
now the speed = sqt (25.16^2+21.9^2)=33.97
i guess the difference between the book answer and this one is sig figs
Your calculations for parts (a) and (c) are correct. However, there is a mistake in your calculation for part (b).
To find the magnitude of the velocity of the rock just before it strikes the ground, you can use the equation:
v^2 = v0^2 + 2ad
where v is the final velocity, v0 is the initial velocity, a is the acceleration, and d is the distance.
In this case, the initial velocity is the magnitude of the horizontal component of the rock's velocity (Vx = 25.16 m/s) and the acceleration is due to gravity (-9.8 m/s^2). The distance (d) is the vertical distance the rock falls from its highest point to the ground, which is 15 m.
Plugging in the values, we have:
v^2 = (25.16)^2 + 2(-9.8)(15)
v^2 = 634.1456 + (-294)
v^2 = 340.1456
Taking the square root of both sides, we get:
v = sqrt(340.1456)
v ≈ 18.44 m/s
Therefore, the magnitude of the velocity of the rock just before it strikes the ground is approximately 18.44 m/s. This differs from the value in the solution book, so it's possible that there is an error in the book or in your original data.
Your approach to parts (a) and (c) of the problem is correct. However, there seems to be a mistake in your calculations for part (b). Let me guide you through the correct calculation.
To find the velocity of the rock just before it strikes the ground, you can first determine the time it takes for the rock to reach the ground from its highest point. This time will be the sum of the time taken to reach the maximum height and the time taken to fall from the maximum height to the ground.
You correctly found the time to reach the maximum height to be 1.6 seconds. Now we need to find the time it takes to fall from the maximum height to the ground.
The equation you used is correct:
-15 = (16.33)t - 4.9t^2
To solve this quadratic equation, bring it to the standard form:
4.9t^2 - 16.33t - 15 = 0
Using the quadratic formula, we have:
t = [-(-16.33) ± √((-16.33)^2 - 4 * 4.9 * (-15))] / (2 * 4.9)
Calculating this, we find two possible values for t: t = 4.58 seconds and t = -0.72 seconds. Since time cannot be negative, we discard the negative value. Therefore, it takes approximately 4.58 seconds for the rock to reach the ground from the highest point.
Now, to find the final velocity:
Vf = Vi + a * t
Vf = 16.33 + (-9.8) * 4.58
Vf = 16.33 - 44.884
Vf = -28.554 m/s
The magnitude of the final velocity is the absolute value of this, which is approximately 28.554 m/s (rounded to 3 decimal places).
Therefore, the correct answer for part (b) is approximately 28.554 m/s, not 22.67 m/s.
Now, let's recap the answers to each part of the problem:
a) The maximum height above the roof reached by the rock: 13.55 m
b) The magnitude of the velocity of the rock just before it strikes the ground: approximately 28.554 m/s
c) The horizontal range from the base of the building to the point where the rock strikes the ground: 100.64 m