A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 57.3° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 21.5 m away. By how much does the rocket clear the top of the wall?
u = 75 cos 57.3
so t = 21.5 /(75 cos 57.3)
Vi = 75 sin 57.3
h = Vi t - 4.9 t^2
answer = h-11
To find out how much the rocket clears the top of the wall, we need to calculate the vertical distance from the rocket's initial position to the top of the wall.
First, split the initial velocity into its horizontal and vertical components.
Vertical component of velocity (V_y) = V * sin(θ)
where V = 75.0 m/s (initial speed of the rocket)
θ = 57.3° (angle above the horizontal)
V_y = 75.0 m/s * sin(57.3°)
V_y ≈ 64.31 m/s
Now, we can find the time it takes for the rocket to reach the top of its trajectory (highest point). At this point, the vertical component of velocity will be zero.
Using the kinematic equation: V_y = V_initial_y + a*t
where V_initial_y = 64.31 m/s (initial vertical velocity)
a = -9.8 m/s^2 (acceleration due to gravity, downward)
t = time
0 = 64.31 m/s + (-9.8 m/s^2) * t
Rearrange the equation to solve for t:
-64.31 m/s = -9.8 m/s^2 * t
t ≈ 6.57 s
Now, we can find the maximum height (h) reached by the rocket:
Using the formula: h = V_initial_y * t + 0.5 * a * t^2
h = 64.31 m/s * 6.57 s + 0.5 * (-9.8 m/s^2) * (6.57 s)^2
h ≈ 212.57 m
The rocket clears the top of the 11.0 m high wall by:
Clearance = h - wall height
Clearance = 212.57 m - 11.0 m
Clearance ≈ 201.57 m
Therefore, the rocket clears the top of the wall by approximately 201.57 meters.
To find out how much the rocket clears the top of the wall, we need to break down the rocket's motion into horizontal and vertical components.
First, we can calculate the time it takes for the rocket to reach the wall by considering only the horizontal motion. We know that the initial speed of the rocket (v0) is 75.0 m/s and the horizontal distance traveled (d) is 21.5 m. Using the formula:
d = v0 * t
where t is the time, we can rearrange the equation to solve for t:
t = d / v0
Substituting the values, we have:
t = 21.5 m / 75.0 m/s
t ≈ 0.287 seconds
Now, let's analyze the vertical motion of the rocket to determine how much it clears the top of the wall. We can use the formula:
h = v0y * t + (0.5) * a * t^2
where h is the maximum height reached by the rocket, v0y is the vertical component of the initial velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).
To find v0y, we use the formula:
v0y = v0 * sin(θ)
where θ is the angle the rocket is fired above the horizontal. Substituting the values, we have:
v0y = 75.0 m/s * sin(57.3°)
v0y ≈ 64.068 m/s
Substituting the values into the equation for h, we have:
11.0 m = (64.068 m/s) * t + (0.5) * (-9.8 m/s^2) * t^2
Rearranging the equation, we get a quadratic equation in terms of t:
0.5 * (-9.8 m/s^2) * t^2 + (64.068 m/s) * t - 11.0 m = 0
Solving this equation, we find two values for t: t1 and t2. We discard the negative value since time cannot be negative in this context.
Using the formula:
t = (-b + √(b^2 - 4ac)) / (2a)
where a = 0.5 * (-9.8 m/s^2), b = (64.068 m/s), and c = -11.0 m, we can solve for t1:
t1 = (-64.068 m/s + √((64.068 m/s)^2 - 4(0.5 * (-9.8 m/s^2) * (-11.0 m)))) / (2 * 0.5 * (-9.8 m/s^2))
Simplifying the equation, we find:
t1 ≈ 2.45 seconds
Therefore, the rocket takes approximately 2.45 seconds to reach its maximum height.
To find out how much the rocket clears the top of the wall, we need to calculate the maximum height it reaches. We can substitute t1 into the equation for h:
h = (64.068 m/s) * (2.45 s) + (0.5) * (-9.8 m/s^2) * (2.45 s)^2
Calculating the value, we find:
h ≈ 46.78 m
Therefore, the rocket clears the top of the wall by approximately 46.78 meters.