Question

An object is thrown with a horizontal velocity (Vh) of 15.0 mt/sec and an initial vertical velocity (Vi) of 29.4 mt/sec. How long will the object take to reach its maximum height and start to return to the earth?
1) 3 Seconds
2) 44.1 meters
3) 6 seconds
4) 90 meters
5) Cannot be determined

Answer 1

Answer:

The object will reach maximum height in approximately $3\; {\rm s}$.

The object will return to the initial height in approximately $6\; {\rm s}$ (assuming that $g = 9.8\; {\rm m\cdot s^{-2}}$, and that the initial vertical velocity points upwards.)

Explanation:

The object in this question is a projectile. If the air resistance on a projectile is negligible, the vertical acceleration of the projectile would be equal to $(-g) = (-9.8)\; {\rm m\cdot s^{-2}}$ during the entire flight.

It is given that the initial velocity in the vertical direction is $29.4\; {\rm m\cdot s^{-1}}$.

• The projectile reaches maximum height when velocity in the vertical direction becomes zero.
• When the projectile returns to the initial position, the vertical velocity would be the opposite of the initial value: $(-29.4)\; {\rm m\cdot s^{-1}}$

To find the duration required, divide the change in velocity (subtract initial velocity from the current value) by acceleration.

By the time the projectile reaches maximum height, vertical velocity would have changed from the initial value to the new value of $0\; {\rm m\cdot s^{-1}}$. Divide the change in velocity by acceleration to find the time required:

$\begin{aligned} & \frac{(\text{new velocity}) - (\text{initial velocity})}{(\text{acceleration})} \\ =\; & \frac{(0) - (29.4)}{(-9.8)} \; {\rm s} \\ \approx \; & 3\; {\rm s{\end{aligned}$.

In other words, the projectile would reach maximum height in approximately $3\; {\rm s}$.

Similarly, by the time the projectile returns to the initial height, vertical velocity would be equal to the opposite of the initial value. The time required would be:

$\begin{aligned} & \frac{(\text{new velocity}) - (\text{initial velocity})}{(\text{acceleration})} \\ =\; & \frac{(-29.4) - (29.4)}{(-9.8)} \; {\rm s} \\ \approx \; & 6\; {\rm s{\end{aligned}$.

In other words, the projectile would return to the initial height in approximately $6\; {\rm s}$.

Answer 2

Final answer:

The object will take approximately 3.96 seconds to reach its maximum height and start to return to the earth.

Explanation:

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 29.4 m/s will spend the same amount of time in the air as a projectile that lands at the same altitude after being thrown straight up with an initial vertical velocity of 29.4 m/s. Therefore, the object will take approximately 3.96 seconds to reach its maximum height and start to return to the earth.