Question

A stone, thrown vertically into the air from ground, returns to the ground in 4s (no air resistance). If the stone is thrown up at twice the initial speed, the time taken to return to the ground will now be what?

Answer 1

When a stone is thrown vertically into the air from the ground, it follows a parabolic trajectory under the influence of gravity. The stone reaches its maximum height and then falls back to the ground. In the absence of air resistance, the time it takes for the stone to return to the ground will be the same as the time it took to reach its maximum height.

Given that the stone returns to the ground in 4 seconds, it means that it takes 2 seconds to reach the maximum height and 2 seconds to descend back to the ground.

Now, if the stone is thrown up at twice the initial speed, the time taken to reach the maximum height will be affected. When an object is thrown vertically upward, the time it takes to reach the maximum height is determined by the initial vertical velocity.

Since the stone is thrown up with twice the initial speed, it means the initial vertical velocity is also doubled. Doubling the initial vertical velocity will affect the time taken to reach the maximum height. The time taken to reach the maximum height can be calculated using the following formula:

$t=\frac{V_f-V_i}{g}$

Where:

• $t$ is the time taken to reach the maximum height,
• $V_f$ is the final vertical velocity (which is 0 at the maximum height),
• $V_i$ is the initial vertical velocity,
• $g$ is the acceleration due to gravity.

In this case, let's denote the original initial velocity as $V_{i1}$ and the doubled initial velocity as $V_{i2}$.

For the original velocity: t1=0−Vi1−9.8=Vi19.8t_1 = frac{0 - V_{i1}}{ -9.8} = frac{V_{i1}}{9.8}t1​=−9.8<span class="si