Question

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Answer 1

Answer: Option (D) 25 m


This is a question of Projectile Motion.

We will use one simple concept: At maximum height, the vertical component of velocity is zero. That is, at max height, only horizontal velocity remains.


We will only use the function of $y$

$y = 10t-t^2 \quad --- (1)$

We need max height. For that, we will need to find the time at which vertical component of velocity becomes zero.

To find vertical component of velocity, we differentiate vertical displacement with respect to time.

We have:

$y=10t-t^2 \\ \\ \\ \implies v_y = \frac{dy}{dt} = 10-2t$

$\text{Now consider }v_y = 0 \\ \\ \implies 10-2t=0 \\ \\ \implies 10=2t \\ \\ \implies t = \frac{10}{2} \\ \\ \implies t = 5 \, \, s$

So, we see that vertical component of velocity becomes zero at $t=5 \, \,s$. We can put t = 5 in (1) to find displacement corresponding to that point, which is the Max Height


$y = 10t-t^2 \\ \\ \text{Max Height reached at }t=5 \\ \\ \implies H_{max} = 10(5)-(5)^2 \\ \\ \implies H_{max} = 50-25 \\ \\ \\ \implies \boxed{H_{max}=25 \, \, m}$

Thus, The Maximum Height Reached is 25 meters.