A ball is thrown vertically upwards. Assuming that air resistance is constant and considerable, how do the time of ascent and the time of descent compare?

Introduction / Context:
In idealised physics problems without air resistance, the motion of a projectile thrown vertically is perfectly symmetric: the time taken to rise to the highest point equals the time taken to fall back to the same level. However, in real life, air resistance often plays an important role. This question asks how constant and considerable air resistance changes the comparison between the time of ascent and the time of descent for a ball thrown straight up.

Given Data / Assumptions:

• A ball is thrown vertically upward from ground level.
• Air resistance is constant and non-negligible throughout the motion.
• Gravity acts downward with acceleration g.
• We compare the time taken to go up to the highest point (ascent) with the time to come back down (descent).

Concept / Approach:
During the ascent, both gravity and air resistance act downward, opposite to the motion. Therefore, the net downward force on the ball is greater than its weight alone, and the magnitude of its acceleration downward is greater than g. This means the ball slows down more quickly and reaches zero velocity in a shorter time, so the ascent is relatively short. During descent, gravity acts downward but air resistance acts upward, opposite to the direction of motion. The net downward force is therefore less than the weight, giving a downward acceleration smaller than g. The ball speeds up more slowly on the way down, so it takes a longer time to return to the starting level. Thus, the time of ascent is less than the time of descent.

Step-by-Step Solution:
Step 1: On the way up, velocity is upward, gravity is downward and air resistance is also downward.Step 2: Net acceleration downward during ascent is greater than g, so the ball loses speed quickly.Step 3: This greater deceleration makes the time of ascent shorter than it would be without air resistance.Step 4: On the way down, velocity is downward, gravity is downward but air resistance is upward.Step 5: Net downward acceleration during descent is less than g, so the ball gains speed more slowly.Step 6: Because the acceleration downward is smaller in magnitude, the time of descent becomes longer.Step 7: Conclude that with significant air resistance, the time of ascent is less than the time of descent.

Verification / Alternative check:
You can think in terms of symmetry breaking. Without air, gravitational acceleration is symmetric on ascent and descent, so the times are equal. Adding air resistance always opposes the motion, making deceleration stronger on the way up and weaker on the way down. This naturally breaks the symmetry and makes the descent take longer. Numerical simulations of projectile motion with drag also confirm that objects typically spend more time descending than ascending for vertical throws.

Why Other Options Are Wrong:
Saying that the time of ascent is greater than the time of descent reverses the correct effect. Claiming that the times are equal is only valid when air resistance is negligible, not when it is constant and considerable as stated in the question. The option none of these is wrong because there is a clear, well-defined relation: ascent time is less than descent time.

Common Pitfalls:
Many students memorise the result that ascent time equals descent time from idealised problems and try to apply it without checking the assumptions. Others may overlook the direction of air resistance relative to motion. To avoid such errors, always examine which forces act at each stage of motion and how they combine to affect acceleration.

Final Answer:
With constant, considerable air resistance, the time of ascent is less than the time of descent.