At the Earth's surface, a projectile is launched straight up at a speed of 9.9 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.
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To determine the height the projectile will rise, we can use the equations of motion.
First, let's convert the initial velocity from km/s to m/s.
9.9 km/s = 9900 m/s
The equation for the maximum height reached by a projectile launched straight up is:
h = (v^2) / (2g)
where h is the maximum height, v is the initial velocity, and g is the acceleration due to gravity near the Earth's surface (which is approximately 9.8 m/s^2).
Plugging in the values, we have:
h = (9900 m/s)^2 / (2 * 9.8 m/s^2)
h = 98010000 m^2/s^2 / 19.6 m/s^2
Simplifying, we get:
h = 5005000 m^2
Therefore, the projectile will rise to a height of approximately 5,005,000 meters.
To calculate the height to which the projectile will rise, you can use the equations of linear motion.
Step 1: Convert the initial velocity from km/s to m/s.
The initial velocity is given as 9.9 km/s. To convert km/s to m/s, multiply by 1000 since there are 1000 meters in a kilometer.
9.9 km/s * 1000 m/km = 9900 m/s
Step 2: Calculate the time taken for the projectile to reach its highest point.
When an object is launched straight up, it will reach its highest point when its vertical velocity becomes zero. The time taken to reach this point can be calculated using the formula:
Time = Final Velocity / Acceleration
In this case, the final velocity is zero since the object is momentarily at rest at its highest point, and the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Time = 0 m/s / 9.8 m/s^2 = 0 seconds
Step 3: Calculate the maximum height reached by the projectile using the formula:
Height = Initial Velocity * Time + (1/2) * Acceleration * Time^2
In this case, the initial velocity is 9900 m/s, the time is 0 seconds, and the acceleration is -9.8 m/s^2 (negative sign because acceleration due to gravity acts in the opposite direction to the initial velocity).
Height = 9900 m/s * 0 s + (1/2) * (-9.8 m/s^2) * (0 s)^2
Since anything multiplied by zero is zero, the height will be zero.
Therefore, the projectile will not rise to any height and will fall back to the Earth immediately after being launched straight up.
(1/2) m v^2 = potential energy
potential energy is mgh for low altitude but that will not work for high altitude where g is less than 9.81 m/s^2
increase in potential = G M m (1/Re - 1/(Re+h)
call z = Re +h
(1/2) v^2 = G M (1/Re - 1/z)
solve for z
then h = z - Re