Josh fires a catapult 18 ft above the ground with an initial velocity of 25 ft/sec. The misslie leaves the catapult at an angle of degree theta with the hoizontal and heads toward a 40 ft wall 500 ft from the launch site. If the missle is fired at a 20 degree angle, does it clear the wall? Justify the answer
If the missle is fired at a 35 degree angle, how high does it go?
The vertical component of the velocity is 25sin20° = 8.55
So, the height y is
y = 18 + 8.55t - 16t^2
If the missile just clears the wall, then
40 = 18 + 8.55t - 16t^2
Hmm. No solutions.
Check for typos, and take a look at wikipedia's article on "trajectory" for details of the range, height, etc.
initial horizontal velocity: 25*cos(theta)=v1
initial vertical velocity:
25*sin(theta)=v2
where theta is the angle to the horizontal.
max height:
(v2)^2-2*32ft/s^2*(height)=0
will it go over the wall:
v1*t=500 ft
solve the above equation for t and plug into:
18+v2*t-(32ft/s^2)/2*t^2= height at given time
one correction:
"max height:
(v2)^2-2*32ft/s^2*(height)=0" is wrong
it should be:
(v2)^2-2*32ft/s^2*(height -18)=0
To determine if the missile clears the wall when fired at a 20-degree angle, we need to calculate its range (horizontal distance traveled).
Let's break down the components of the initial velocity into horizontal and vertical directions:
- The horizontal component is calculated as V₀x = V₀ * cos(theta), where V₀ is the initial velocity (25 ft/sec) and theta is the launch angle (20 degrees).
- The vertical component is calculated as V₀y = V₀ * sin(theta).
Using these components, we can calculate the time it takes for the missile to reach its peak height above the ground (when it has maximal vertical velocity) in the following way:
- For a projectile launched vertically, the time to reach the peak height is t_peak = V₀y / g, where g is the acceleration due to gravity (32.2 ft/sec²).
Next, we can calculate the time it takes for the missile to reach the ground (total flight time) by doubling the time to reach the peak height:
- t_total = 2 * t_peak
Using the total flight time, we can calculate the range (horizontal distance) the missile travels:
- range = V₀x * t_total
To determine if the missile clears the 40 ft wall, we compare the calculated range to the distance of the wall (500 ft):
- If the range is greater than 500 ft, the missile clears the wall. Otherwise, it does not.
Now, let's apply the calculations:
- V₀x = 25 ft/sec * cos(20 degrees) ≈ 23.20 ft/sec
- V₀y = 25 ft/sec * sin(20 degrees) ≈ 8.54 ft/sec
- t_peak = 8.54 ft/sec / 32.2 ft/sec² ≈ 0.265 seconds
- t_total = 2 * 0.265 s ≈ 0.53 seconds
- range = 23.20 ft/sec * 0.53 s ≈ 12.30 ft
Since the calculated range of approximately 12.30 ft is less than the distance of the wall (500 ft), the missile does not clear the wall when fired at a 20-degree angle.
To calculate the maximum height reached by the missile when fired at a 35-degree angle, we can use the same components of the initial velocity:
- V₀x = 25 ft/sec * cos(35 degrees)
- V₀y = 25 ft/sec * sin(35 degrees)
The time to reach the peak height (t_peak) and the total flight time (t_total) can be calculated using the same formulas as for the 20-degree angle case above.
The maximum height reached by the missile is given by:
- height = V₀y² / (2 * g)
Now, let's apply the calculations:
- V₀x = 25 ft/sec * cos(35 degrees) ≈ 20.41 ft/sec
- V₀y = 25 ft/sec * sin(35 degrees) ≈ 14.33 ft/sec
- t_peak = 14.33 ft/sec / 32.2 ft/sec² ≈ 0.445 seconds
- t_total = 2 * 0.445 s ≈ 0.89 seconds
- height = (14.33 ft/sec)² / (2 * 32.2 ft/sec²) ≈ 6.38 ft
Therefore, when fired at a 35-degree angle, the missile reaches a maximum height of approximately 6.38 ft above the ground.