Analysis and Calculation of Time to Reach the Ground for an Object Thrown Vertically Upwards

When an object is thrown vertically upwards with an initial velocity, its journey is influenced by the force of gravity. This article will explore how to calculate the total time it takes for the object to reach the ground, using both analytical methods and the equations of motion.

Initial Parameters and Equations of Motion

Consider an object thrown vertically upwards with an initial velocity of 14 m/s. The acceleration due to gravity is -9.81 m/s2. Using the equations of motion, we can solve for the time it takes for the object to return to the ground.

Initial parameters:

Initial velocity (u): 14 m/s Acceleration due to gravity (g): -9.81 m/s2 (direction is downwards) Final displacement (s): 0 m (ground level)

The displacement equation of motion is given by:

s ut 0.5at^2 s ut 0.5at^2

Substituting the known values:

0 14t 0.5(-9.81)t^2 0 14t 0.5(-9.81)t^2

This simplifies to:

0 14t - 4.905t^2 0 14t - 4.905t^2

Factoring out t:

t(14 - 4.905t) 0 t(14 - 4.905t) 0

Therefore, the solutions are:

t 0 (initial time when the object is thrown) 14 - 4.905t 0 which simplifies to t ≈ 2.85 seconds

Thus, the time taken to reach the ground is approximately 2.85 seconds.

Alternative Method of Calculation

Another method involves considering the symmetry of the object's motion. In its upward journey, the object moves against gravity, and its speed decreases until it reaches zero velocity at the highest point. Once it stops, it starts moving downward under the influence of gravity, covering the same distance in the same amount of time as it took to reach the highest point.

The time to reach the highest point is given by:

t frac{u}{g} frac{20}{9.81} ≈ 2.04 seconds t frac{u}{g} frac{20}{9.81} ≈ 2.04 seconds

Since the motion is symmetrical, the time to fall back to the ground is also 2.04 seconds. Therefore, the total time to reach the ground is:

2.04 seconds 2.04 seconds 4.08 seconds

Coordinate System and Mathematical Formulation

To solve this problem, we can set the point of projection on the ground as the origin of the coordinate system. The upward direction is considered positive, and the downward direction is considered negative.

The equation to find the time to reach the ground when the object is projected upwards with an initial velocity of 20 m/s is:

s ut frac{1}{2}at^2 s ut frac{1}{2}at^2

Substituting the known values:

0 20t - 10t^2 0 20t - 10t^2

This simplifies to:

0 20t - 5t^2 0 20t - 5t^2

Factoring out t:

t(20 - 5t) 0 t(20 - 5t) 0

Therefore, the solutions are:

t 0 (initial time when the object is thrown) 20 - 5t 0 which simplifies to t 4 seconds

Thus, the object will return to the ground at t 4 seconds.