A projectile is being launched from ground level with no air resistance. You want to avoid having it enter a temperature inversion layer in the atmosphere a height h above the ground.
1) What is the maximum launch speed you could give this projectile if you shot it straight up? Express your answer in terms of h and g .
2)Suppose the launcher available shoots projectiles at twice the maximum launch speed you found in part A. At what maximum angle above the horizontal should you launch the projectile?
3)How far (in terms of h) from the launcher does the projectile in part B land?
1) v=sqrt(2gh)
2) sqrt(2gh)/2sqrt(2gh)= 1/2
therefore arsin(1/2)= 30'
1) To avoid having the projectile enter the temperature inversion layer at height h, we can use the equations of motion for vertically upward motion:
The maximum height reached by the projectile can be found using the equation:
h_max = (v^2)/(2g)
where v is the initial velocity (launch speed) and g is the acceleration due to gravity. In this case, the projectile is shot straight up, so the final velocity at the highest point is zero (v_final = 0).
So, we have:
0 = v^2 - 2gh_max
Solving for v^2:
v^2 = 2gh_max
Therefore, the maximum launch speed is:
v = √(2gh_max)
2) If the launcher shoots projectiles at twice the maximum launch speed found in part A, the new maximum launch speed would be 2v, where v is the maximum launch speed from part A.
To find the maximum angle above the horizontal at which to launch the projectile, we can use the equation for range:
R = (v^2 * sin(2θ))/g
where R is the range, θ is the launch angle, and all other variables have their usual meanings.
We want to find the launch angle that gives the maximum range. Using calculus, we can find that the maximum range occurs when sin(2θ) = 1. Therefore:
1 = sin(2θ)
2θ = 90 degrees
θ = 45 degrees
So, the maximum angle above the horizontal at which to launch the projectile is 45 degrees.
3) The range of the projectile is given by the equation:
R = (v^2 * sin(2θ))/g
In part B, the maximum angle above the horizontal at which the projectile is launched is 45 degrees, and the maximum launch speed is 2v, where v is the maximum launch speed in part A.
Substituting these values into the equation, we have:
R = [(2v)^2 * sin(2*(45 degrees))]/g
The sin(90 degrees) term simplifies to 1, so we have:
R = [(2v)^2 * 1]/g
R = (4v^2)/g
Therefore, the distance the projectile lands from the launcher is 4 times the maximum height from part A, in terms of h.
To answer these questions, we need to use kinematic equations and principles of projectile motion. Here's how we can approach each part:
1) To find the maximum launch speed for the projectile, we can consider the time it takes for the projectile to reach its maximum height and start falling back down. When the projectile reaches its peak, its vertical velocity becomes zero. Using the kinematic equation for vertical motion, we have:
v = u + (-g)t
Where:
v = final vertical velocity (zero when reaching the peak)
u = initial vertical velocity (upward since it is shot straight up)
g = acceleration due to gravity (9.8 m/s²)
t = time taken to reach the peak
Initially, the projectile is at rest, so u = 0. On reaching the peak, the vertical velocity becomes zero, so v = 0. Solving for t, we have:
0 = 0 + (-9.8)t
t = 0 seconds
However, this gives us an incorrect answer, as the projectile takes some time to reach the peak (t > 0). We need to solve the equation again but this time taking into account the time for the projectile to reach the peak. The total time to reach the peak is given by:
t_total = 2t
Since the projectile reaches the peak at half of the total time. Now we can solve for the maximum launch speed (v_max):
h = u(t_total) + (1/2)(-g)(t_total)²
Since we want to avoid the temperature inversion layer, the projectile must not reach a height h. So, h is the maximum height reached by the projectile. Therefore:
h = (1/2)(-g)(t_total)²
h = (-4.9)(t_total)²
Now, solving for t_total:
t_total² = -2h / -4.9
t_total = √(2h / 4.9)
Finally, substituting the value of t_total into the equation for velocity:
v_max = u + (-g)(t_total)
v_max = -9.8 * √(2h / 4.9)
So, the maximum launch speed you could give the projectile when shot straight up is -9.8 * √(2h / 4.9) m/s.
2) If the launcher shoots projectiles at twice the maximum launch speed found in part 1, then the new launch speed is 2 * (-9.8 * √(2h / 4.9)) = -19.6 * √(2h / 4.9).
To find the maximum angle above the horizontal at which to launch the projectile, we need to split the launch velocity into horizontal and vertical components. Let's denote the launch angle as θ.
Horizontal component of velocity (v_x) = (launch speed) * cos(θ)
Vertical component of velocity (v_y) = (launch speed) * sin(θ)
Since we want to determine the maximum launch angle, we can assume the projectile will land at the same horizontal distance as the launcher. This means the range of the projectile is equal to the distance from the launcher to the landing point.
The range (R) of a projectile can be calculated using the range formula:
R = (launch speed) * v_x * (2 * sin(θ) / g)
Substituting the previous values, we have:
R = (-19.6 * √(2h / 4.9)) * cos(θ) * (2 * sin(θ) / 9.8)
To find the maximum angle, we differentiate R with respect to θ, set it equal to zero, and solve for θ:
dR/dθ = 0
Solving this equation will give us the maximum angle above the horizontal at which to launch the projectile.
3) To determine how far the projectile in part B lands from the launcher, we need to find the horizontal distance traveled (range, R). We can use the range formula again:
R = (launch speed) * v_x * (2 * sin(θ) / g)
Substituting the launch speed from part B, we have:
R = (-19.6 * √(2h / 4.9)) * cos(θ) * (2 * sin(θ) / 9.8)
Now, we need to solve for R in terms of h.