Question

A baseball is hit straight up into the air at a speed of $20 \mathrm{~m} / \mathrm{s}$. How long does it take to reach the high-point in its trajectory?

Answer 1

Hint: In a neat table of units, list the value of each $(s, u, v, a, t)$ factor. The factor you are searching for gets a question mark and a dash gets the irrelevant one. Decide your equation and rearrange it so that the subject is the factor you are looking for. Units to solve and add. Projectile motion is the motion, subject only to the acceleration of gravity, of an object thrown or projected into the air.

Formula used:
$a=\dfrac{v-u}{t}$

Complete answer:
Projectile motion is when, in a bilaterally symmetrical, parabolic path, an object moves. Its trajectory is called the path that the object follows. We'll see the definition of projectile motion in this question first. Then we'll see the horizontal and vertical component expression of the particle velocity that is projected from the ground at the angle $\theta$. Then the velocity and kinetic energy are finally calculated at the highest point.
Step 1
$s=-m$
$u=20 \dfrac{m}{s}$
$v=0 \dfrac{m}{s}$
$a=-9.81 \mathrm{~ms}^{-2} .$
$t=? s$

Step 2
$a=\dfrac{v-u}{t}$ so $t=\dfrac{v-u}{a}$

Step 3
$t=\dfrac{0-20}{-9.81}$
$\therefore t=2.04s$

It $2.04s$ take to reach the high-point in its trajectory.

Note:
The object is referred to as a projectile, and its trajectory is called its path. Projectile motion only happens when at the beginning there is one force applied, after which gravity is the only influence on the trajectory. The time taken by a projectile to cover the whole trajectory is referred to as flight time. Descent Time: The time taken by the body to reach the trajectory from maximum height to the lowest level is called the descent time. This is an expression of the maximum height that the projectile reaches.