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  • In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m s^–1 at 45° from a height 1.5 m above ground.

In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m s-1 at 45^{\circ} from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s-2, the kinetic energy of the shotput when it just reaches the ground will be
(a) 2.5 J
(b) 5.0 J
(c) 52.5 J
(d) 155.0 J

Answers (1)

The answer is the option (d) 155.0 J

h = 1.5 m, v = 1 m/s, m = 10 kg, g = 10 ms-2

the initial energy of shot put can be calculated as:

PE + KE = mgh + \frac{1}{2} mv^{2}

= 10 \times 10 \times 1.5 + \frac{1}{2} \times 10 \times 1 = 155 J

From the law of conservation of energy, we can write that,

(PE + KE) initial = (PE + KE) final

Hence, the final kinetic energy of the shotput: KE final + 0 = 155J

Hence the answer is 155 J

Posted by

infoexpert24

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