Question

A baseball is thrown straight up at $15m{{s}^{-1}}$. How high will it go?

Answer 1

Hint: In this question, we have given a condition where a ball is thrown up vertically and we need to find the maximum height achieved by it. Since, the ball is going upward this means that it will keep going in upward direction until its velocity will become here. Here the acceleration will act in the opposite direction of motion and such type of acceleration is known as retardation.

Complete answer:
We have been given that the speed of the base ball is $15m{{s}^{-1}}$
Therefore, we can say that initial Velocity (u) = $15m{{s}^{-1}}$
Now, since the ball is going against gravity, the acceleration will be negative.
Therefore, we can say that acceleration (a) = - (g)
Where “g” is the acceleration due to gravity.
Also, we need to find that the maximum height it will achieve and for this condition we know that we will be having the final velocity as zero
Therefore, v = 0
Let the distance travelled be “x”
Now, we have all the required things
So now by using equation of laws of motion
${{v}^{2}}-{{u}^{2}}=2as$
Putting the values, we get
$0-{{(15)}^{2}}=2(-9.8)(s)$
$s=11.479$meters
Therefore, we can say that the ball will achieve a height of 11.479 meters.

Note:
Since, the acceleration is acting in the opposite direction of motion, this means that the acceleration is reducing the speed of the ball which is thrown vertically upward and hence we will take acceleration as negative. Also keep in mind that when a ball is thrown straight upwards then the formula of maximum height from parabola concept will not be valid as the ball is moving in a straight line and not in the form of parabola.