If a rocket is shot vertically upward from the ground with an initial velocity of 192 ft/sec.
a. When does it reach its maximum height above the ground?
b. What is the maximum height reached by the rocket?
c. How long does it take to reach the ground again?
d. At what speed will the rocket hit the ground?
the height h is given by
h = 192t - 16t^2
(a,b) find the vertex of the parabola
(c) solve for t when h=0
(d) final speed is the same as the starting speed, since momentum is conserved
To answer these questions, we can use the equations of motion for vertical motion of an object under gravity. The key equation is:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration due to gravity (approximately -32 ft/sec^2)
t = time
a. To find when the rocket reaches its maximum height, we need to find the time at which the rocket's velocity becomes zero.
0 = 192 - 32t
Solving this equation gives us:
t = 192 / 32 = 6 seconds
Therefore, the rocket reaches its maximum height 6 seconds after it is shot.
b. To find the maximum height reached by the rocket, we can use the equation for displacement:
s = ut + (1/2)at^2
Substituting the values, we get:
s = 192 * 6 + (1/2) * (-32) * (6^2)
s = 1152 - 576 = 576 feet
Hence, the maximum height reached by the rocket is 576 feet.
c. To find the time it takes for the rocket to reach the ground again, we need to find the time taken from the maximum height to the ground. Since the motion is symmetrical, this time will be the same as the time taken to reach the maximum height.
Therefore, it takes 6 seconds for the rocket to reach the ground again.
d. Finally, to find the speed at which the rocket hits the ground, we can use the equation:
v = u + at
where u is the final velocity (zero in this case) and a is the acceleration due to gravity (-32 ft/sec^2).
Substituting the values, we get:
v = 0 + (-32) * 6
v = -192 ft/sec
The negative sign indicates that the final velocity is downward, which is expected since the rocket is falling towards the ground. Therefore, the rocket hits the ground with a speed of 192 ft/sec.
To answer these questions, we can use the kinematic equations of motion. The key equations we need are:
1. v = u + at
2. s = ut + 0.5at^2
3. v^2 = u^2 + 2as
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is time
- s is the displacement
Note: In this scenario, we are assuming negligible air resistance.
a. When does the rocket reach its maximum height above the ground?
To find this, we need to determine the time it takes for the vertical velocity to become zero when the rocket is moving upward. We can use equation 1.
Given:
u = 192 ft/sec
a = -32 ft/s^2 (assuming acceleration due to gravity is -32 ft/s^2)
Let's calculate the time it takes to reach maximum height (t1).
Since the rocket is moving vertically upward, the acceleration due to gravity acts in the opposite direction, so we take its negative value.
At maximum height, the final vertical velocity (v1) is zero.
Using equation 1:
v1 = u + at1
0 = 192 - 32t1
32t1 = 192
t1 = 192 / 32
t1 = 6 seconds
Therefore, the rocket reaches its maximum height after 6 seconds.
b. What is the maximum height reached by the rocket?
To find the maximum height, we can use equation 2.
Given:
u = 192 ft/sec
a = -32 ft/s^2
t = 6 seconds
Let's calculate the maximum height (s1).
Using equation 2:
s1 = ut1 + 0.5a(t1)^2
s1 = 192 * 6 + 0.5 * (-32) * (6)^2
s1 = 1152 - 576
s1 = 576 ft
Therefore, the maximum height reached by the rocket is 576 feet.
c. How long does it take to reach the ground again?
To find the time it takes to reach the ground, we need to determine when the displacement becomes zero. Again, we can use equation 2.
Given:
u = 192 ft/sec
a = -32 ft/s^2
Let's calculate the time it takes to reach the ground again (t2).
Using equation 2:
0 = ut2 + 0.5a(t2)^2
0 = 192t2 - 16(t2)^2
16(t2)^2 - 192t2 = 0
16t2(t2 - 12) = 0
t2 = 0 (initial time) or t2 - 12 = 0
t2 = 0 (not feasible in this scenario) or t2 = 12 seconds
Therefore, it takes 12 seconds for the rocket to reach the ground again.
d. At what speed will the rocket hit the ground?
To find the speed at which the rocket hits the ground, we can use equation 1.
Given:
u = 192 ft/sec
a = -32 ft/s^2
t = 12 seconds
Let's calculate the final velocity (v2) when the rocket hits the ground.
Using equation 1:
v2 = u + at2
v2 = 192 - 32 * 12
v2 = 192 - 384
v2 = -192 ft/sec (negative sign indicates downward velocity)
Therefore, the rocket hits the ground with a speed of 192 ft/sec in the downward direction.