Response:
The ball remained airborne for 3.896 seconds
Explanation:
Given that
g = 9.8 m/s², representing gravitational acceleration,
If the angle of launch is 45°, the horizontal range will be maximized.
Both horizontal and vertical launch velocities are equal, each equating to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The duration to reach maximum height is half of the flight time.
v = u + at ∵ v = 0 (at maximum height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The total time in the air equals twice the time to reach maximum height
2 t₁ = 3.896 s
The horizontal distance covered is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in the air for 3.896 seconds
The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)
For your instance, Vhorizontal = 16 · cos(40) equates to 12.3 m/s
Conclusion: 12.3 m/s
Answer:
Explanation:
As the baseball ascends, gravitational forces as well as air resistance act downward, whereas the displacement is moving upward which results in an angle of 180° between the force and displacement. Therefore, the work done by both the gravitational force and air resistance is negative, confirming option (d) as accurate.
