The highest barrier that a projectile can clear is 18.8 m, when the projectile is launched at an angle of 59.0 ° above the horizontal. What is the projectile's launch speed?
do vertical problem with conservation of energy first:
(1/2) m Vi^2 = m g h
.5 Vi^2 = 9.81 * 18.8
Vi = 19.2 m/s initial speed up
sin 59 = 19.2 / s
s = 22.4 m/s
To find the projectile's launch speed, we can use the principles of projectile motion and trigonometry. The key information we have from the problem is the maximum height the projectile can reach, which is 18.8 m, and the launch angle, which is 59.0°.
The projectile's trajectory can be split into vertical and horizontal components. The horizontal component represents the projectile's motion along the x-axis, and the vertical component represents the projectile's motion along the y-axis.
First, let's find the time it takes for the projectile to reach its maximum height. At the highest point of its trajectory, the vertical component of the projectile's motion is zero. We can use the vertical motion equation:
vf = vi + a * t
where vf is the final vertical velocity (which is zero at the highest point), vi is the initial vertical velocity (which we need to find), a is the vertical acceleration (which is gravitational acceleration, approximately -9.8 m/s^2), and t is the time.
Since we know that vf = 0, we can rearrange the equation to solve for vi:
0 = vi - 9.8 * t
Solving for vi:
vi = 9.8 * t
Next, we need to find the time it takes for the projectile to reach its maximum height. At the highest point, the vertical velocity is zero, so we can use the following equation to calculate the time:
vf = vi + a * t
where vf is the final vertical velocity (which is zero at the highest point), vi is the initial vertical velocity (which we just found), a is the vertical acceleration (which is gravitational acceleration, approximately -9.8 m/s^2), and t is the time.
Since we know that vf = 0, we can rearrange the equation to solve for t:
0 = vi - 9.8 * t
Substituting the value of vi with 9.8 * t:
0 = 9.8 * t - 9.8 * t
0 = 0
This means that the time it takes for the projectile to reach its maximum height is zero. This might seem counterintuitive, but it makes sense because at the highest point of its trajectory, the vertical velocity is momentarily zero before the projectile starts to come back down. Therefore, the time it takes for the projectile to reach its maximum height is negligible.
Now, let's consider the horizontal component of the projectile's motion. We can use the horizontal equation of motion:
x = vi * t
where x is the horizontal displacement, vi is the initial horizontal velocity (which we need to find), and t is the time.
Since we know that the projectile's horizontal displacement is zero (because it returns to its initial position after its maximum height), we can rearrange the equation to solve for vi:
0 = vi * t
From this equation, we can see that either the initial horizontal velocity (vi) or the time (t) must be zero. However, since we know that the time it takes to reach the maximum height is negligible, we can conclude that the initial horizontal velocity (vi) is zero.
Finally, we can use trigonometry to find the projectile's launch speed. The launch speed is the magnitude of the initial velocity vector, which is the combination of the horizontal and vertical components of the velocity. We have already found that the initial horizontal velocity is zero, so the launch speed is equal to the initial vertical velocity (vi).
Therefore, the launch speed of the projectile is 9.8 m/s.