An arrow is shot into the air. Its range is 2 | vedclass

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(B) Given:Range $R = 200\,m$Time of flight $T = 5\,s$Acceleration due to gravity $g = 10\,m/s^2$$1$. Horizontal component of velocity $(u_x)$:The hori

Vedclass · Accepted Answer

$40\,m/s, 31.25\,m$

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(B) Given:Range $R = 200\,m$Time of flight $T = 5\,s$Acceleration due to gravity $g = 10\,m/s^2$$1$. Horizontal component of velocity $(u_x)$:The horizontal range is given by $R = u_x 	imes T$.Therefore,$u_x = \frac{R}{T} = \frac{200}{5} = 40\,m/s$.$2$. Maximum height $(H)$:The time of flight is given by $T = \frac{2u_y}{g}$,where $u_y$ is the vertical component of velocity.$5 = \frac{2u_y}{10} \implies 2u_y = 50 \implies u_y = 25\,m/s$.The maximum height is given by $H = \frac{u_y^2}{2g}$.$H = \frac{(25)^2}{2 	imes 10} = \frac{625}{20} = 31.25\,m$.Thus,the horizontal component of velocity is $40\,m/s$ and the maximum height is $31.25\,m$.

An arrow is shot into the air. Its range is $200\,m$ and its time of flight is $5\,s$. If $g=10\,m/s^2$,then the horizontal component of velocity and the maximum height will be respectively:

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