A projectile is fired such that the vertical component of its velocity is 49 m/s. The horizontal component of its velocity is 60 m/s.
a) how long does the projectile remain in the air?
b) what horizontal distance does it travel?
Now we have done a lot of these. You try.
To find the answers, we can break down the problem into its horizontal and vertical components.
a) To determine how long the projectile remains in the air, we need to find the time it takes for the projectile to reach its highest point and then come back down to the ground. This can be done by considering the vertical motion of the projectile.
We know the initial vertical velocity is 49 m/s, and the acceleration due to gravity is approximately 9.8 m/s² (assuming no air resistance).
Using the equation of motion: final velocity (vf) = initial velocity (vi) + (acceleration x time)
At the highest point, the final velocity (vf) will be zero since the projectile momentarily stops before falling downward.
Using vf = 0, vi = 49 m/s, and acceleration = -9.8 m/s² (negative sign indicates downward direction), we can rearrange the equation to solve for time (t):
0 = 49 - (9.8 x t)
9.8t = 49
t = 49 / 9.8
t = 5 seconds
Therefore, the projectile remains in the air for 5 seconds.
b) To determine the horizontal distance traveled by the projectile, we can use the horizontal component of its velocity.
We know the horizontal velocity is 60 m/s, and the time of flight is 5 seconds (from part a).
The distance traveled can be calculated using the formula: distance = velocity x time.
distance = 60 m/s * 5 s
distance = 300 meters
Hence, the horizontal distance traveled by the projectile is 300 meters.