Answer:
The ball was in air for 3.896 s
Explanation:
given,
g = 9.8 m/s², acceleration due to gravity,
If the launch angle is 45°, the horizontal range will be maximum.
The horizontal and vertical launch velocities are equal, and each is equal to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The time to attain maximum height is one half of the time of flight.
v = u + at ∵ v = 0 (max. height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The time of flight is twice of the maximum height time
2 t₁ = 3.896 s
The horizontal distance traveled is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in air for 3.896 s
