A projectile thrown from the ground level at an angle of alpha, is vertically displaced 35 meters within the last one second before landing.How many meters might it have raised ? (Assume ;g=10 N/kg,There is no drag)

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A projectile thrown from the ground level at an angle of alpha, is vertically displaced 35 meters within the last one second before landing.How many meters might it have raised ? (Assume ;g=10 N/kg,There is no drag)

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Answer 1 · 2016-04-25T10:04:04

$80 m$ $80m$

Explanation:

As understood the problem.

Let $h$ $h$ be the height from which the projectile starts its downward flight. Also let $t$ $t$ be the time of downward flight. Let this point be origin.
As drag is to be ignored,
Kinematic equation is
$s = u t + \frac{1}{2} g t^{2}$ $s=ut+1/2g t^2$ , inserting given values
$h_{t} = \frac{1}{2} \times 10 t^{2} = 5 t^{2}$ $h_t=1/2xx10 t^2=5t^2$ .....(1)

Let us find distance of flight in time $t - 1$ $t-1$
$h_{t - 1} = 5 {(t - 1)}^{2}$ $h_(t-1)=5(t-1)^2$ .....(2)

It is given that displacement in $t^{t h}$ $t^(th)$ second is 35 meters. Subtracting (2) from (1) and setting it equal to the given value.
$h_{t} - h_{t - 1} = 5 t^{2} - 5 {(t - 1)}^{2} = 35$ $h_t-h_(t-1)=5t^2-5(t-1)^2=35$ , Simplifying and solving for $t$ $t$

$5 t^{2} - 5 (t^{2} - 2 t + 1) = 35$ $5t^2-5(t^2-2t+1)=35$
or $(10 t - 5) = 35$ $(10t-5)=35$
$\Rightarrow t = 4 s$ $=>t=4s$

From (1)
$h_{t} = 5 \times 4^{2}$ $h_t=5xx4^2$
$= 80 m$ $=80m$

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A projectile thrown from the ground level at an angle of alpha, is vertically displaced 35 meters within the last one second before landing.How many meters might it have raised ? (Assume ;g=10 N/kg,There is no drag)

1 Answer

Explanation:

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