A projectile is shot vertically upward with a given initial velocity. It reaches a maximum height of 100 m. If, on a second shot, the initial velocity is doubled then the projectile will reach a maximum height of ____________?
initially, all the energy is KE = 1/2 mv^2
At the peak, all the KE has become PE = mgh
So, with 2v at first, there's 4 times the KE. That means 4 times the height at the peak.
To find the maximum height of the projectile when the initial velocity is doubled, we need to first understand the relationship between the initial velocity and the maximum height.
The maximum height reached by a vertically upward fired projectile can be calculated using the kinematic equation:
vf^2 = vi^2 + 2aΔy
Where:
vf = final velocity (when the projectile reaches its maximum height, the final velocity is zero)
vi = initial velocity
a = acceleration (in this case, acceleration due to gravity, which is approximately -9.8 m/s^2)
Δy = change in vertical position or the maximum height
Since the final velocity is zero at the maximum height, the equation becomes:
0 = vi^2 + 2aΔy
Rearranging the equation, we get:
Δy = -vi^2 / (2a)
We are given that the maximum height is 100 m for the first shot. Let's denote it as Δy1 and the initial velocity for the first shot as vi1. Then we can write the equation as:
100 = -vi1^2 / (2a) ---(1)
Now, for the second shot, where the initial velocity is doubled, let's denote the maximum height as Δy2 and the initial velocity for the second shot as vi2. The equation becomes:
Δy2 = -vi2^2 / (2a) ---(2)
We want to find Δy2.
Comparing equations (1) and (2), we can see that the only difference is in the initial velocities:
100 = -vi1^2 / (2a)
Δy2 = -vi2^2 / (2a)
Since the initial velocity is doubled for the second shot (vi2 = 2 * vi1), we can substitute this into equation 2:
Δy2 = -((2 * vi1)^2) / (2a)
Δy2 = -4 * (vi1^2) / (2a)
Δy2 = -2 * (vi1^2) / a
Since we know that Δy1 = 100, we can substitute it into equation 1:
100 = -vi1^2 / (2a)
Rearranging the equation, we can express vi1^2 in terms of a:
vi1^2 = -200a
Substituting back into equation 2:
Δy2 = -2 * (-200a) / a
Δy2 = 400
Therefore, the maximum height reached by the projectile when the initial velocity is doubled is 400 m.
To calculate the height reached by a projectile when the initial velocity is doubled, we can use the following equation for the maximum height (H) reached by a projectile:
H = (V₀²) / (2g)
Where:
H = Maximum height reached
V₀ = Initial velocity
g = Acceleration due to gravity (approximately 9.8 m/s²)
In the given scenario, the initial velocity is doubled, so we can write the new initial velocity as 2V₀ (where V₀ is the initial velocity of the first shot).
Let's substitute these values into the equation:
H = ((2V₀)²) / (2g)
H = (4V₀²) / (2g)
H = (2V₀²) / g
We know that the maximum height reached in the first shot is 100 m. So we can substitute H = 100 m and solve for the new initial velocity:
100 = (2V₀²) / 9.8
To find the new maximum height, we need to calculate the new initial velocity (2V₀) using the equation above and then substitute it into the original equation for the maximum height.
Solving for V₀:
100 × 9.8 = 2V₀²
V₀² = (100 × 9.8) / 2
V₀² = 490
V₀ = √490
V₀ ≈ 22.14 m/s
Now we can substitute the new initial velocity into the original equation to find the new maximum height (H):
H = (2V₀²) / g
H = (2 × (22.14 m/s)²) / 9.8 m/s²
H ≈ 97.3 m
Therefore, when the initial velocity is doubled, the projectile will reach a maximum height of approximately 97.3 meters.