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
a. 13.6m
b. 34.6m/s
c. 103m
b) solution is like this.
Y=vyt-1/2gt^2 --------------(i)
15=16.33t-5t^2 (g=10m/s)
Use quadratic method.find t
The value of t comes 4.01s...
Again use euqation (i)
To find vy ,where t=4.01s.
Then vy comes 23.79m/s.
Now,
for magnitude.
V^2=(25.16)^2+(23.79)^2
: .V=34.6m/s
At first to find time use (-15).
Then to find vy use just (+15).
Kyle,
Thanks for solving thw first part
V=√vx^2+vy^2
a) Well, you've got this one right! The maximum height reached by the rock is indeed 13.55 meters above the roof.
b) Now, let's see what went wrong here. You correctly set up the equation -15 = (16.33)t - 4.9t^2. However, when solving for t, you seem to have made an error. The correct solution should be t = 2.71 seconds. Plugging this back into the equation Vf = Vi + at, we get Vf = 16.33 - 9.8(2.71) = -11.28 m/s. Taking the magnitude of this gives us a final velocity of approximately 11.28 m/s, not 22.67 m/s.
c) Finally, to find the horizontal range, you can use the formula x = Vx*t. Plugging in the values, we get x = 25.16 * 2.71 = 68.34 meters. So, the correct answer for the horizontal range is approximately 68.34 meters, not 100.64 meters.
Remember, math can be tough sometimes, but don't worry! Mistakes happen to the best of us. Keep up the good work and keep clowning around with those calculations!
Let's go through each part of the problem:
a) To find the maximum height reached by the rock, you correctly found the initial x and y components of the velocity. However, the value of Vy you calculated is incorrect. It should be Vy = 30*sin(33°) = 15.98 m/s.
You correctly determined the time it took for the rock to reach its highest point, which is 1.6 seconds. To find the maximum height above the roof, you need to use the equation for vertical displacement:
H_max = Vy*t - (1/2)*g*t^2
Plugging in the values, we get:
H_max = (15.98 m/s)(1.6 s) - (1/2)(9.8 m/s^2)(1.6 s)^2
H_max = 25.57 m
So, the maximum height above the roof reached by the rock is 25.57 meters.
b) To find the magnitude of the velocity of the rock just before it strikes the ground, you correctly determined the time it takes for the rock to fall. However, there is an error in the equation setup. It should be:
-15 m = 15.98 m/s * t - (1/2)*(9.8 m/s^2)*t^2
Simplifying and rearranging, we get:
4.9 t^2 - 15.98 t - 15 = 0
Using the quadratic formula, we find two solutions for t: t = 0.652 s and t = 4.097 s. Since we're interested in the time it takes to fall, we discard the smaller value, t = 0.652 s.
Now we can find the final velocity of the rock using the equation:
Vf = Vy - g*t
Vf = 15.98 m/s - (9.8 m/s^2)(4.097 s)
Vf = -20.66 m/s
Taking the absolute value, the magnitude of the velocity just before it strikes the ground is 20.66 m/s (not -22.67 m/s).
c) You correctly used the time of 3.98 s to find the horizontal range. The equation for horizontal displacement is:
x = Vx * t
Plugging in the values, we get:
x = 25.16 m/s * 3.98 s
x = 100.12 m
So, the horizontal range from the base of the building to the point where the rock strikes the ground is approximately 100.12 meters (not 100.64 m).
To summarize the correct answers:
a) The maximum height above the roof reached by the rock is 25.57 meters.
b) The magnitude of the velocity of the rock just before it strikes the ground is 20.66 m/s.
c) The horizontal range from the base of the building to the point where the rock strikes the ground is 100.12 meters.
Magnitude is not the absolute value of the velocity. Magnitude is the combination of velocity in the x and y directions. You have to use Pythagorean Theron to solve for magnitude.
For this problem it would be...
Vf^2=22.6^2 + 30^2