Question

a ball is thrown upwards at a speed of 15 m/s. how long does it take the ball to reach the highest point

Answer 1

Approximately
`1.5\; {\rm s}` assuming that air resistance is negligible and
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Under the assumptions, the ball would be accelerating at a constant
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` in the vertical direction during the whole flight.

Note that the value of
`a` is negative because the ball is accelerating towards the ground. While the ball was initially travelling upwards away from the ground (at
`u = 15\; {\rm m\cdot s^(-1)}`,) the ball would slow down, reaches a velocity of zero, and travel towards the ground as the velocity of the ball becomes negative.

When the ball is travelling upwards, the height of the ball increases. However, as soon as the velocity of the ball becomes negative, the ball would start returning to the ground. Hence, the height of the ball is maximized right before the velocity of the ball becomes negative, which is when the velocity of the ball is at
`0\; {\rm m\cdot s^(-1)}`.

The time required to reach that point of maximum height can be found by dividing the change in velocity by acceleration:

• Initial velocity:
  `u = 15\; {\rm m\cdot s^(-1)}`.
• Velocity when height is maximized:
  `v = 0\; {\rm m\cdot s^(-1)}`.
• Acceleration (rate of change in velocity):
  `a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`.

Time required:

`\begin{aligned}t &= (v - u)/(a) \\ &= \frac{0\; {\rm m\cdot s^(-1)} - 15\; {\rm m\cdot s^(-1)}}{(-9.81)\; {\rm m\cdot s^(-2)}} \\ &\approx 1.5\; {\rm s}\end{aligned}`.

In other words, the ball would reach the highest point in approximately
`1.5\; {\rm s}`.