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Velocity of COM

The horizonta...

Question

The horizontal and vertical displacement x and y of a projectile at a given time t are given by x= $6 t$ metre and y= $8 t$ - $5 t^{2}$ metre. The range of the projectile in metre is.

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Solution

$Y = 8 t - 5 t^{2}$

From this equation

$u S i n θ = 8$

$x = 6 t$

From this equation

$u C o s θ = 6$

Square and add equations (1) and (2)

$u ² (S i n ² θ + C o s ² θ) = 8 ² + 6 ²$

$u ² = 100$

$u = 10 m / s$

Velocity of projection is $10 m / s$

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Similar questions

Q. The horizontal and vertical displacements

x

y

of a projectile at given time

t

x = 6 t (m)

y = 8 t - 5 t^{2} (m)

. The range of the projectile in metres is :

y

and the distance

x

along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by

y = (8 t - 5 t^{2})

x = 6 t

t

is in seconds. The velocity of projection is:

y

and the distance

x

(both in m) along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by

y = (8 t - 5 t^{2})

x = 6 t

t

is in seconds. The velocity of projection is

y

and horizontal distance

x

covered by a projectile in a time

t

seconds are given by the equations

y = 8 t - {5 t}^{2}

x = 6 t

x

y

are measured in metres, the velocity of projection is:

y

and the distance

x

along the horizontal plane of a projectile on a certain planet (assuming flat surface) with no surrounding atmosphere are given by

x = 6 t

y = 8 t - 5 t^{2}

x

y

are in metre and time

t

is in second. Find the velocity in

m s^{- 1}

with which the body is projected. Take acceleration due to gravity

g = 10 {m/s}^{2}

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