A bullet is fired vertically upwards with an initial velocity of 50 m $s^{- 1}$ . It covers a distance h, during the first second and a distance $h_{2}$ during the last 3 seconds of its upward motion. If g = 10 m $s^{- 2}$ , how are $h_{1}$ and $h_{2}$ related to each other.

h_{1} = h_{2}

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h_{1} = 2 h_{2}

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2 h_{1} = h_{2}

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3 h_{1} = h_{2}

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Solution

The correct option is A $h_{1} = h_{2}$
using equation of motion : $s = u t + \frac{a t^{2}}{2}$
distance traveled in first second:
$h_{1} = 45 m$
$t_{a} = \frac{u}{g}$ ,where $t_{a}$ is time of ascent
$t_{a} = 5 s$
so at $t = 2 s$ last 3 sec of upward motion starts
height at $t = 2 s$ is :
$h = 50 \times 2 - \frac{10 \times 4}{2}$
$h = 80 m$
$H_{m a x} = \frac{u^{2}}{2 g}$
$H m a x = 125 m$
distance traveled in last 3 sec $h_{2} = H - h$
$h_{2} = 125 - 80$
$h_{2} = 45 m$
therefore $h_{1} = h_{2}$

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A bullet is fired vertically upwards with an initial velocity of 50 m s−1. It covers a distance h, during the first second and a distance h2 during the last 3 seconds of its upward motion. If g = 10 m s−2, how are h1 and h2 related to each other.

A bullet is fired vertically upwards with an initial velocity of 50 m $s^{- 1}$ . It covers a distance h, during the first second and a distance $h_{2}$ during the last 3 seconds of its upward motion. If g = 10 m $s^{- 2}$ , how are $h_{1}$ and $h_{2}$ related to each other.