Consider the motion of a projectile. It isfired at t = 0. Its initial speed is 36 m/s and its initial projection angle is 55◦ from the horizontal.
The acceleration of gravity is 9.8 m/s2 . y
x
What is the maximum height, h, of its tra jectory?
Answer in units of m.
Maximum height is achieved when the velocity's vertical component is zero.
36*sin55 - gt = 0
Solve for the time t. Then use
y = 30*sin55* t - (g/2)t^2
to get the maximum altitude h.
To find the maximum height of the projectile's trajectory, we can use the equations of motion and kinematics.
First, let's focus on the vertical component of the motion. We can break down the initial velocity of the projectile into its vertical and horizontal components. The initial velocity in the vertical direction can be found using trigonometry:
Vertical component of velocity (Vy) = Initial speed × sin(initial projection angle)
Vy = 36 m/s × sin(55°)
Vy ≈ 29.06 m/s (rounded to two decimal places)
Now, let's consider the vertical motion of the projectile. At the highest point of its trajectory, the vertical component of velocity becomes zero. We can use the equation of motion in the vertical direction to find the time taken to reach the maximum height:
Final velocity in the vertical direction (Vy') = 0 m/s
Acceleration in the vertical direction (ay) = -9.8 m/s^2 (negative sign because the acceleration is in the opposite direction of motion)
Using the equation: Vy' = Vy + (ay × t)
0 m/s = 29.06 m/s - (9.8 m/s^2 × t)
Solving for time (t):
9.8 m/s^2 × t = 29.06 m/s
t = 29.06 m/s / 9.8 m/s^2
t ≈ 2.97 seconds (rounded to two decimal places)
Now that we know the time taken to reach the maximum height, we can find the maximum height (h) using the equation of motion in the vertical direction:
h = Vy × t + (1/2)ay × t^2
h = 29.06 m/s × 2.97 s + (1/2)(-9.8 m/s^2)(2.97 s)^2
h ≈ 43.98 m (rounded to two decimal places)
Therefore, the maximum height of the projectile's trajectory is approximately 43.98 meters.