Question

A cricket ball is hit vertically upwards and returns to ground 6s later. Calculate (i) maximum height reached by the ball (ii) initial velocity of a ball

(45 m, 30ms-¹)​

Answer 1

Answer:

45 m, 30 m/s

Explanation:

The equation of motions are

$v=u+gt\\\\v^{2} = u^{2} + 2gs\\ \\s = ut + \frac{1}{2}gt^{2}$

t= 6 s

so, we can say that at time t = 3s, the ball reached the maximum height.

therefore, $v = 0 m/s$

we need to find u

value of g = -10 m/s² because the motion is against gravity

$v = u + gt\\\\0 = u + (-10)(3)\\\\0= u-30\\30 = u\\\\u = 30 m/s$

so initial velocity is 30 m/s.

now,

$v^{2} =u^{2} + 2gs\\\\0 = 900 + 2(-10)(s)\\\\20s = 900\\\\s=\frac{900}{20} \\\\s=45 m$

Ans maximum height is 45 m

Answer 2

Answer:

(1) maximum height = 45 m

(2) initial velocity = 30m/s

Explanation:

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