A projectile is shot at 35° to the horizontal and lands 12 s later at the same elevation.
(a) What is the projectile's initial speed?
(b) What is its maximum altitude?
To solve this problem, we can use the equations of motion for projectile motion and the known information provided. Here's how we can solve it:
(a) To determine the initial speed of the projectile, we can first split the initial velocity into its horizontal and vertical components.
Let's consider the horizontal component first. The initial velocity in the horizontal direction remains constant throughout the motion. We can use the equation:
Horizontal distance = Horizontal velocity × Time
Since the horizontal velocity remains constant and time is given as 12 seconds, we can write:
Horizontal distance = Initial horizontal velocity × 12 seconds
However, in this case, the horizontal distance is not given. Instead, it is implied that the projectile lands at the same elevation as where it was shot. Therefore, the horizontal distance is zero.
Substituting the values, we get:
0 = Initial horizontal velocity × 12 seconds
Since the horizontal distance is zero, the initial horizontal velocity is also zero.
Now let's consider the vertical component of the initial velocity. We know that the motion is in a parabolic shape, and the equation for the vertical displacement is:
Vertical displacement = (Initial vertical velocity × Time) + (0.5 × Gravity × Time^2)
Since the projectile lands at the same elevation as where it was shot, the vertical displacement is also zero. We can write:
0 = (Initial vertical velocity × 12 seconds) + (0.5 × Gravity × (12 seconds)^2)
Simplifying this equation, we get:
0 = Initial vertical velocity × 12 seconds + 6 seconds^2 × Gravity
Now, we also know that the initial velocity can be expressed as the sum of its vertical and horizontal components:
Initial velocity = √(Initial horizontal velocity^2 + Initial vertical velocity^2)
Since we found earlier that the initial horizontal velocity is zero, the equation simplifies to:
Initial velocity = √(Initial vertical velocity^2)
Substituting the equation for initial vertical velocity from the previous step, we get:
Initial velocity = √((-6 seconds^2 × Gravity) / 12 seconds)
Simplifying further, we have:
Initial velocity = √((-0.5 × Gravity × Time^2) / Time)
Since the time cancels out, the equation becomes:
Initial velocity = √(-0.5 × Gravity × Time)
Plugging in the values, with Gravity being 9.8 m/s^2 and Time being 12 seconds, we can calculate the initial velocity.
(b) To find the maximum altitude, we can use the equation for the vertical displacement again. However, in this case, we're interested in the highest point of the trajectory, where the vertical velocity is momentarily zero.
Using the equation:
Vertical displacement = (Initial vertical velocity × Time) + (0.5 × Gravity × Time^2)
We can substitute the values we know:
Vertical displacement = (Initial vertical velocity × 12 seconds) + (0.5 × Gravity × (12 seconds)^2)
Simplifying and rearranging this equation, we have:
Initial vertical velocity × 12 seconds = -0.5 × Gravity × (12 seconds)^2
Initial vertical velocity = (-0.5 × Gravity × (12 seconds)^2) / (12 seconds)
Plugging in the values for Gravity and Time, we can calculate the initial vertical velocity.
Once we have the initial vertical velocity, we can find the highest point of the trajectory by dividing it by the acceleration due to gravity (9.8 m/s^2) to get the time it takes to reach the maximum altitude. Then we can substitute this time into the equation for vertical displacement to find the maximum altitude.