A projectile is shot horizontally at 34.5m/s from the edge of a 55.0metre building
a) Determine the height of the projectile from the ground after 2.0 seconds.
b) Calculate the velocity of the projectile when the projectile reaches the ground.
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To solve this problem, we can use the equations of motion for projectile motion. Here's how we can find the solution step by step:
a) To find the height of the projectile from the ground after 2.0 seconds, we need to determine the time it takes for the projectile to reach the ground. Since the projectile is shot horizontally, its initial vertical velocity is zero.
Step 1: Determine the time of flight.
Using the equation for vertical displacement of an object in free fall:
h = ut + (1/2)gt^2,
where h is the vertical displacement (height), u is the initial vertical velocity (0 m/s in this case), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
In this case, the vertical displacement is equal to the height of the building: h = 55.0 m.
We can rearrange the equation to solve for time:
t^2 = (2h / g),
t = sqrt(2h / g).
Substituting the known values:
t = sqrt(2 * 55.0 m / 9.8 m/s^2),
t ≈ sqrt(11.22 s^2),
t ≈ 3.35 s (approximately).
The time of flight is approximately 3.35 seconds.
Step 2: Determine the height of the projectile after 2.0 seconds.
Since the time given is less than the time of flight, we can use the equation:
h = ut + (1/2)gt^2,
where u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s^2), t is the time (2.0 s).
Substituting the known values:
h = 0 + (0.5 * 9.8 m/s^2 * (2.0 s)^2),
h = 0 + (0.5 * 9.8 m/s^2 * 4.0 s^2),
h ≈ 19.6 meters (approximately).
Therefore, the height of the projectile from the ground after 2.0 seconds is approximately 19.6 meters.
b) To calculate the velocity of the projectile when it reaches the ground, we need to use the equation for horizontal motion.
Since there is no horizontal acceleration, the horizontal velocity remains constant at the initial velocity throughout the motion. Therefore, the projectile's horizontal velocity is still 34.5 m/s.
The time to reach the ground is the same as the time of flight, which we found to be approximately 3.35 seconds.
Using the equation for horizontal displacement:
d = vt,
where d is the horizontal displacement, v is the horizontal velocity, and t is the time.
Substituting the known values:
d = 34.5 m/s * 3.35 s,
d ≈ 115.57 meters (approximately).
Therefore, the velocity of the projectile when it reaches the ground is approximately 115.57 m/s.