A ball is thrown upwards

A ball is thrown upwards. Neglecting air resistance, what initial upward speed does the ball need to remain in the air for a total time of 10 seconds?

about 60 m/s
about 50 m/s
about 100 m/s
about 110 m/s
about 80 m/s

The correct answer and explanation is :

To determine the initial speed required for a ball to stay in the air for a total time of 10 seconds, we use kinematic equations. The key concept here is that the ball’s motion consists of two equal parts:

  1. Ascending phase (from launch to the highest point).
  2. Descending phase (from the highest point back to the ground).

Step 1: Understanding Time of Flight

The total time in the air is 10 seconds, meaning the ball takes 5 seconds to reach its peak and another 5 seconds to fall back down.

Step 2: Using the Kinematic Equation

For an object moving under gravity, the time to reach the highest point is given by:

[
t = \frac{v_0}{g}
]

where:

  • ( v_0 ) is the initial velocity (m/s),
  • ( g ) is the acceleration due to gravity (9.81 m/s²),
  • ( t ) is the time to reach the highest point (5 seconds).

Rearranging the equation for ( v_0 ):

[
v_0 = g \cdot t
]

Substituting the values:

[
v_0 = (9.81 \, \text{m/s}^2) \times (5 \, \text{s})
]

[
v_0 \approx 49.05 \, \text{m/s}
]

Step 3: Choosing the Correct Answer

Among the options given, the closest value is about 50 m/s.

Conclusion

The correct answer is: about 50 m/s.

This means that if a ball is thrown straight up with an initial speed of about 50 m/s, it will reach its highest point in 5 seconds and return to the ground in another 5 seconds, staying in the air for a total of 10 seconds.

Scroll to Top