A) A projectile is fired straight upward at
143 m/s.
How fast is it moving at the instant it
reaches the top of its trajectory?
Answer in units of m/s.
B)
How fast is it moving at the instant it reaches
the top of its trajectory if the projectile is
fired upward at 12◦
from the horizontal?
Answer in units of m/s.
A) To find the speed of the projectile at the instant it reaches the top of its trajectory, we can use the principle of conservation of energy. At the topmost point of its trajectory, the projectile's kinetic energy will be zero since it momentarily comes to a stop. Therefore, all its initial energy is in the form of potential energy.
The potential energy of an object at a height h is given by the equation: potential energy = mass x acceleration due to gravity x height.
Since the projectile is fired straight upward, its height at the topmost point is the maximum height reached during its trajectory. At this point, the potential energy is equal to the initial energy, which is the initial kinetic energy of the projectile.
The initial kinetic energy of the projectile is given by the equation: kinetic energy = 0.5 x mass x velocity^2.
Equating the two expressions for energy, we can solve for the velocity at the top:
0.5 x mass x velocity^2 = mass x acceleration due to gravity x height
Canceling out the mass:
0.5 x velocity^2 = acceleration due to gravity x height
Solving for velocity:
velocity = sqrt(2 x acceleration due to gravity x height)
In this case, the height is the maximum height reached by the projectile. Since the projectile is fired straight upward, it eventually stops and comes back down. Therefore, the maximum height reached is when the projectile's vertical velocity component becomes zero.
Using the formula for vertical velocity (v = u + at), where initial vertical velocity (u) is 143 m/s, acceleration (a) is -9.8 m/s^2 (taking downwards as negative), and solving for time (t):
0 = 143 - 9.8t
t = 143/9.8
Now, using the formula for displacement (s = ut + 0.5at^2), where displacement (s) is the maximum height reached by the projectile, initial vertical velocity (u) is 143 m/s, acceleration (a) is -9.8 m/s^2, and solving for displacement (s):
s = 143(143/9.8) + 0.5(-9.8)(143/9.8)^2
Once we determine the maximum height reached, we can substitute this value into the formula for velocity to calculate the speed at the top of the trajectory using the equation:
velocity = sqrt(2 x acceleration due to gravity x maximum height reached)
B) To find the speed of the projectile at the instant it reaches the top of its trajectory when it is fired upward at an angle, we need to break down its initial velocity into horizontal and vertical components. The vertical component will determine the maximum height reached, and the horizontal component will not affect the vertical motion.
Given that the projectile is fired upward at 12◦ from the horizontal, we can find the vertical and horizontal components of the initial velocity using trigonometry.
Vertical component (v_y) = initial velocity x sin(angle)
Horizontal component (v_x) = initial velocity x cos(angle)
Using the vertical component, we can follow the same process as in part A to find the maximum height reached by the projectile. Once we have the maximum height, we can use the formula for velocity (explained in part A) to find the speed at the top of the trajectory.
Keep in mind that the angle of projection does not affect the answer, as long as the vertical component is used to determine the maximum height.