A small rocket is fired in a test range. It rises high into the air and soon runs out of fuel. On the way down it passes near an observer (sitting in a 21.5-m-high tower) who sees the rocket traveling at a speed of 35 m/s and moving in a vertical plane at an angle of 56° below horizontal.
1. What is the maximum altitude reached?
Vertical speed is= -29.016
Find how far it fall to reach-> -29.016
v=-gt
-29.016=-9.81t
t= 2.961s
Find distance it fell in that time:
d=1/2gt^2
d=4.9(2.961)^2
d=42.961
From its peak height to the height of the observer if fell 42.961m then add the height of the tower (21.5m) this will give max altitude. 64.461m
B. with what speed does the rocket hit the ground?
64.461=1/2gt^2
t= 13.155
from here I am stuck. Any help please.
I'm sorry to hear that you're stuck, but don't worry, I'm here to help! To find the speed at which the rocket hits the ground, we can use the time it takes for the rocket to reach the ground after running out of fuel.
Since we know the rocket's initial vertical velocity is -29.016 m/s (negative because it's moving downward), and the time it takes to reach the ground is 13.155 seconds, we can use the equation:
v = u + gt
where:
v = final velocity (which we want to find)
u = initial velocity (which is -29.016 m/s)
g = acceleration due to gravity (-9.81 m/s^2)
t = time taken (which is 13.155 seconds)
Plugging in the values, we have:
v = -29.016 + (-9.81)(13.155)
Simplifying this equation will give us the speed at which the rocket hits the ground. I'll let you do the math and find out the answer! Let me know if you have any other questions or need further assistance.
To find the speed at which the rocket hits the ground, we need to calculate the horizontal distance it travels before reaching the ground.
We know that the rocket's initial vertical speed is -29.016 m/s and it will take approximately 13.155 seconds to hit the ground.
Using these values, we can calculate the total vertical distance traveled during this time:
d = Vinitial * t + 1/2 * g * t^2
d = -29.016 * 13.155 + 1/2 * (-9.81) * (13.155)^2
d ≈ -381.428 + (-78.596) = -460.024 m
Note that the negative sign indicates the downward direction.
Now, we need to find the horizontal distance traveled during this time. We can use the horizontal component of the velocity:
v_horizontal = v * cos(angle)
v_horizontal = 35 * cos(56°)
v_horizontal ≈ 35 * 0.5592 ≈ 19.5736 m/s
Finally, we can calculate the horizontal distance using the equation:
d_horizontal = v_horizontal * t
d_horizontal ≈ 19.5736 * 13.155 ≈ 257.207 m
Since the horizontal distance is positive, it means the rocket traveled horizontally in the positive direction.
The speed at which the rocket hits the ground can be found using the horizontal and vertical distances:
Speed = sqrt(d_horizontal^2 + d^2)
Speed = sqrt(257.207^2 + (-460.024)^2)
Speed ≈ sqrt(66172.160 + 211648.205) ≈ sqrt(277820.365) ≈ 527.087 m/s
So, the rocket hits the ground with a speed of approximately 527.087 m/s.