a projectile is fired horizontally at 13.4m/s from the edge of a 9.50m high cliff and strikes the ground.
(a) what is the horizontal distance it traveled.
(b)the elapsed time
(c)the final velocity
To find answers to these questions, we can use the equations of motion for horizontal and vertical motion separately. Let's solve each part step by step:
(a) Horizontal distance traveled:
Since the projectile is fired horizontally, the initial horizontal velocity is constant throughout the motion. Therefore, the horizontal distance traveled is equal to the horizontal velocity multiplied by the time of flight.
To calculate the time of flight, we can use the equation for vertical motion. When the projectile hits the ground, its vertical displacement will be equal to the height of the cliff (9.50m), and the initial vertical velocity is zero. Using these values, we can find the time it takes for the projectile to reach the ground.
The equation for vertical displacement is:
y = ut + (1/2)gt^2,
where:
y = vertical displacement (9.50m),
u = initial vertical velocity (0),
g = acceleration due to gravity (-9.8 m/s^2),
t = time of flight.
Plugging in the known values and solving for t:
9.50 = (1/2)(-9.8)t^2
19 = -9.8t^2
t^2 = 19 / -9.8
t ≈ 2.11 seconds (taking the positive value since time cannot be negative)
Now that we have the time of flight, we can calculate the horizontal distance traveled:
Distance = Horizontal Velocity × Time of Flight
Distance = 13.4 m/s × 2.11 s ≈ 28.274 meters
So, the horizontal distance traveled is approximately 28.274 meters.
(b) Elapsed time:
The elapsed time is the same as the time of flight, which we calculated in part (a). Therefore, the elapsed time is approximately 2.11 seconds.
(c) Final velocity:
In the horizontal direction, the velocity remains constant throughout the projectile's motion. Therefore, the final horizontal velocity is the same as the initial horizontal velocity, which is 13.4 m/s.
In the vertical direction, the final velocity can be calculated using the equation:
v = u + gt,
where:
v = final vertical velocity (unknown),
u = initial vertical velocity (0),
g = acceleration due to gravity (-9.8 m/s^2),
t = time of flight (2.11 seconds).
Plugging in the values:
v = 0 + (-9.8)(2.11)
v ≈ -20.678 m/s (taking the negative value as the final velocity points downwards)
So, the final velocity is approximately -20.678 m/s.