a projectile is fired vertically upward from ground level with a velocity of 1600 ft/sec. If air resistance is neglected, find its distance s(t) above ground at time t. What is its maximum height?
s(t) = -16t^2 + 1600 t
the time of max height is on the axis on symmetry
... t = -1600 / (2 * -16)
the slope of the tangent (1st deriv) is equal to zero at max height
Why did the projectile join the circus? Because it wanted to reach new heights, just like you!
Now, let's calculate the distance above the ground (s(t)) at any given time (t) for our flying friend. We can use the equation:
s(t) = s(0) + v(0)*t - (1/2) * g * t^2
where:
s(t) is the distance above the ground at time t,
s(0) is the initial position (ground level in this case),
v(0) is the initial velocity (1600 ft/sec),
g is the acceleration due to gravity (32 ft/sec^2), and
t is the time.
Plugging in the values we have:
s(t) = 0 + (1600 ft/sec) * t - (1/2) * (32 ft/sec^2) * t^2
Now, to find the maximum height, we need to find the time when the projectile reaches its highest point. At the maximum height, the vertical velocity becomes zero because the projectile starts falling back down. We can find this time by using this equation:
v(t) = v(0) - g * t
At the maximum height, v(t) is zero, so:
0 = (1600 ft/sec) - (32 ft/sec^2) * t_max
Solving for t_max:
t_max = (1600 ft/sec) / (32 ft/sec^2)
t_max = 50 seconds
Now, let's find the maximum height by substituting t_max into the equation for s(t):
s_max = 0 + (1600 ft/sec) * (50 seconds) - (1/2) * (32 ft/sec^2) * (50 seconds)^2
And after some calculations:
s_max = 4000 feet
So, the projectile's maximum height is 4000 feet. Keep reaching for the stars (or at least 4000 feet)!
To find the distance above the ground at time t and the maximum height, we need to use the equations of motion for a projectile under constant acceleration. In this case, the only force acting on the projectile is gravity, so the acceleration is -32 ft/sec² (negative because it acts downward).
Let's break down the problem step by step:
Step 1: Find the equation for the distance above the ground at time t.
The equation for the vertical position of the projectile as a function of time is given by:
s(t) = s₀ + v₀t + (1/2)at²
In this equation:
- s(t) is the distance above the ground at time t.
- s₀ is the initial position (ground level), which is 0 in this case.
- v₀ is the initial velocity, which is 1600 ft/sec.
- a is the acceleration, which is -32 ft/sec².
- t is the time.
Substituting the known values into the equation, we get:
s(t) = 0 + 1600t - 16t²
So the equation for the distance above the ground at time t is s(t) = 1600t - 16t².
Step 2: Find the maximum height.
The maximum height occurs at the vertex of the parabolic path, which corresponds to the maximum value of s(t).
To find the maximum value of s(t), we need to determine the vertex of the parabola. The vertex of a parabola is given by the equation:
t = -b / (2a)
In our equation s(t) = 1600t - 16t², a = -16 and b = 1600.
Substituting the values, we get:
t = -1600 / (2 * -16)
t = 50 seconds
To find the maximum height, substitute this value of t back into the equation for s(t):
s(t) = 1600t - 16t²
s(50) = 1600 * 50 - 16 * (50)²
Calculating this, we get:
s(50) = 80000 - 16 * 2500
s(50) = 80000 - 40000
s(50) = 40000 ft
Hence, the maximum height reached by the projectile is 40000 ft.
To find the distance above the ground at time 't' and the maximum height reached by the projectile, we can use the equations of motion for vertical motion under constant acceleration. Let's break down the problem step by step:
Step 1: Find the equation for the position of the projectile, s(t), as a function of time.
The equation for the position of an object undergoing vertical motion is given by:
s(t) = s_0 + v_0t + (1/2)gt^2
Where:
s(t) is the position at time t
s_0 is the initial position (ground level)
v_0 is the initial velocity (upward velocity in this case)
g is the acceleration due to gravity (-32.2 ft/sec^2, taking into account the downward direction)
In this case, the initial position s_0 is at ground level, so s_0 = 0.
Therefore, the equation for the position becomes:
s(t) = v_0t - (1/2)gt^2
Step 2: Determine the initial velocity, v_0.
In this problem, it is given that the initial velocity is 1600 ft/sec in the upward direction.
v_0 = 1600 ft/sec
Step 3: Calculate the maximum height reached by the projectile.
To find the maximum height, we need to find the time at which the projectile has a zero vertical velocity. At the maximum height, the object momentarily stops moving upward and starts falling downward.
When the projectile reaches its maximum height, its vertical velocity becomes zero (v = 0).
Using the equation of motion for vertical velocity, v(t), which is the derivative of the position equation with respect to time:
v(t) = v_0 - gt
We can substitute v = 0 to find the time t when the velocity is zero:
0 = v_0 - gt
Solving for t:
t = v_0/g
Substituting the given values:
t = 1600 ft/sec / 32.2 ft/sec^2
Step 4: Substitute the time t into the equation for the position of the projectile.
Using the equation for the position from Step 1:
s(t) = v_0t - (1/2)gt^2
Substituting the values of v_0 and t:
s(t) = (1600 ft/sec) * (1600 ft/sec / 32.2 ft/sec^2) - (1/2) * 32.2 ft/sec^2 * (1600 ft/sec / 32.2 ft/sec^2)^2
Step 5: Simplify the equation for the position.
s(t) = 1600t - 16.1t^2
Step 6: Calculate the maximum height.
To find the maximum height, we substitute the time t from Step 3 into the position equation.
s_max = 1600t - 16.1t^2
Substituting the value of t:
s_max = 1600 ft/sec * (1600 ft/sec / 32.2 ft/sec^2) - 16.1 ft/sec^2 * (1600 ft/sec / 32.2 ft/sec^2)^2
Simplifying this expression will give you the maximum height reached by the projectile.
By evaluating the equation, the maximum height can be calculated.