By adding the time of flight at the top of the cliff and the time taken down the hill, the ball spent 2.72 s of time in air.
<h3>
What is a Projectile ?</h3>
A projectile can be a stone or a ball or any object that can be projected. The object projected will take a trajectory path.
Given that acceleration due to gravity is 9.8 m/s and a 0.87 kg projectile is fired into the air from the top of a 8.52 m cliff above a valley. Its initial velocity is 9.1 m/s at 49◦ above the horizontal. To know how long the projectile will be in the air, we will calculate the total time of flight at the top of the cliff and the time taken down the hill
Time of flight T = 2usinФ/g
Where
Substitute all the necessary parameters into the formula
T = (2 × 9.1sin49)/ 9.8
T = 13.74/9.8
T = 1.40 s
The time taken down the cliff can be found with formula
h = ut + 1/2gt²
where u = 0
8.52 = 0 + 1/2 × 9.8 × t²
8.52 = 4.9t²
t² = 8.52/4.9
t² = 1.738
t = √1.738
t = 1.32 s
Time for the projectile in the air = T + t
Time in air = 1.40 + 1.32
Time = 2.72 s
Therefore, the projectile will be in the air for 2.72 seconds long.
Learn more about Projectile here: brainly.com/question/24949996
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