the height h is
h(x) = x tanθ - g/(2(v cosθ)^2) x^2
= .7x - .0008 x^2
h(32) = 21.58
So, it clears the wall by 21.6-15.2 = 6.4m
h(x) = x tanθ - g/(2(v cosθ)^2) x^2
= .7x - .0008 x^2
h(32) = 21.58
So, it clears the wall by 21.6-15.2 = 6.4m
First, let's find the time it takes for the rocket to reach the wall using the horizontal component of its motion. The horizontal distance traveled by the rocket can be found using the formula: distance = speed * time. Since the speed of the rocket is 96.0 m/s and the distance to the wall is 32.0 m, we can rearrange the formula to solve for time: time = distance / speed.
time = 32.0 m / 96.0 m/s = 0.333 s
Next, let's find the maximum height reached by the rocket using the vertical component of its motion. The vertical distance traveled by the rocket can be found using the formula: distance = initial velocity * time + (1/2) * acceleration * time^2. Since the rocket starts at ground level, the initial vertical velocity is 0 m/s. The acceleration due to gravity is -9.8 m/s^2 (negative sign because it acts downwards). The time we found earlier is the time it takes for the rocket to reach the wall, so we can use this time in the formula.
distance = 0 * 0.333 s + (1/2) * (-9.8 m/s^2) * (0.333 s)^2
distance = -0.541 m (vertical distance below ground level)
Since the rocket starts from ground level and the vertical distance is negative, we can conclude that the rocket clears the top of the wall by:
Clearance = wall height - (vertical distance + height)
Clearance = 15.2 m - (-0.541 m + 0 m)
Clearance = 15.2 m + 0.541 m
Clearance ≈ 15.741 m
Therefore, the rocket clears the top of the wall by approximately 15.741 m.