A projectile is shot straight up from the earth's surface at a speed of 1.20×10^4 km/h.
How high does it go?
To find how high the projectile goes, we need to use the equations of projectile motion. First, let's convert the initial speed of the projectile from km/h to m/s, as the standard unit for speed in physics is meters per second.
Given: Initial speed = 1.20 × 10^4 km/h
1 km = 1000 m
1 hour = 3600 seconds
To convert km/h to m/s, we can use the conversion factor:
1 km/h = (1000 m) / (3600 s)
Therefore:
1.20 × 10^4 km/h = (1.20 × 10^4 * 1000 m) / (3600 s)
Simplifying, we get:
1.20 × 10^4 km/h = 3.33 × 10^2 m/s
Now that we have converted the initial speed to m/s, we can proceed to determine the maximum height reached by the projectile.
The motion of the projectile can be divided into two phases: upward motion and downward motion. During the upward motion, the only force acting on it is gravity, which causes the projectile to slow down and eventually come to a stop before reversing direction. At the highest point, the velocity becomes momentarily zero before the downward motion begins.
To determine the maximum height reached, we can use the following equation of projectile motion:
v_f^2 = v_i^2 + 2aΔy
where:
- v_f is the final velocity (which becomes zero at the maximum height)
- v_i is the initial velocity
- a is the acceleration due to gravity (~9.8 m/s²)
- Δy is the change in vertical displacement (maximum height)
Since the final velocity is zero at the maximum height, the equation becomes:
0 = (3.33 × 10^2 m/s)^2 + 2(9.8 m/s²)Δy
Solving for Δy (the maximum height), we have:
2(9.8 m/s²)Δy = - (3.33 × 10^2 m/s)^2
Δy = - (3.33 × 10^2 m/s)^2 / (2(9.8 m/s²))
Simplifying and calculating, we find:
Δy ≈ -5.64 × 10^3 m²
However, it's important to note that the minus sign in Δy indicates that it refers to the vertical displacement from the starting point, which is downward. To find the actual height, we need to take the absolute value of Δy:
Height = | Δy | = 5.64 × 10^3 m
Therefore, the projectile reaches a height of approximately 5.64 kilometers.
To find the height the projectile reaches, we need to follow these steps:
Step 1: Convert the initial velocity from km/h to m/s.
Step 2: Calculate the time it takes for the projectile to reach its highest point.
Step 3: Use the time obtained in step 2 to find the maximum height using the formula for vertical displacement.
Let's calculate step by step:
Step 1: Convert the initial velocity from km/h to m/s.
Given, initial velocity (V) = 1.20×10^4 km/h.
To convert km/h to m/s, we can use the conversion factor: 1 km/h = 0.2778 m/s.
So, V = 1.20×10^4 × 0.2778 m/s = 3333 m/s.
Step 2: Calculate the time it takes for the projectile to reach its highest point.
To find the time, we can use the formula for vertical motion:
Vf = Vi + gt,
where Vf is the final velocity, Vi is the initial velocity, g is the acceleration due to gravity, and t is the time.
At its highest point, the final velocity becomes zero. Therefore, Vf = 0.
0 = 3333 m/s - 9.8 m/s^2 × t.
Simplifying, we get:
t = 3333 m/s ÷ 9.8 m/s^2 = 339.69 s.
Step 3: Use the time obtained in step 2 to find the maximum height using the formula for vertical displacement.
The formula for vertical displacement is:
Δy = Vi × t + 0.5 × g × t^2.
Given Vi = 3333 m/s, g = 9.8 m/s^2, and t = 339.69 s.
Let's calculate:
Δy = 3333 m/s × 339.69 s - 0.5 × 9.8 m/s^2 × (339.69 s)^2.
Simplifying, we find:
Δy = 5.66 × 10^6 m.
Therefore, the projectile goes to a maximum height of 5.66 × 10^6 meters.
Since that is quite a high launch velocity, comparable to the escape velocity, you have to take into account the decrease in the acceleration of gravity with distance. You can do that by using the correct formula for the potential energy function V(r), where r is the distance from the center of the Earth. At launch, r1 = re = 6378*10^3 m.
The initial velocity in m/s is
V1 = 3333 m/s.
At the highest altitude, V2 = 0
The potential energy can be written
V(r) = - M g'/r
where g' is the value at the earth's surface (re), 9.80 m/s^2
Solve this equation for the largest distance from earth center, r2:
-Mg'[(1/r) - (1/re)] = (1/2) M V1^2
The M's cancel out. Solve for r and subtract re for the altitude.