Question

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Answer 1

Hey There!!

Here a ball is thrown vertically upwards. It comes back in 6 seconds. 

Since the nature of motion of ball is symmetric, it means that it takes 3 seconds to reach the highest point.

Let us suppose the ball was thrown with a velocity u. At the highest point, its velocity becomes zero. Also, gravitational acceleration g is acting downwards, opposite to direction of motion.


From equations of motion, we have:

$v=u+at \\ \\ \\ \implies 0 = u + (-9.8)(3) \\ \\ \\ \implies u = 9.8 \times 3 \\ \\ \\ \implies \boxed{u=29.4 \, \, m / s}$

Thus, Initial Velocity of the Ball was 29.4 m/s



Now, we also have to find the maximum height. We use another equation of motion.



Let maximum height be h.



$v^2=u^2+2as \\ \\ \\ \implies 0 = (29.4)^2 + 2(-9.8)h \\ \\ \\ \implies h = \frac{29.4 \times 29.4}{2 \times 9.8} \\ \\ \\ \implies \boxed{h=44.1 \, \, m}$


Thus, Maximum Height that the Ball reached was 44.1 metres




Hope it helps
Purva
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