A ball is projected with a horizontal velocity of $5.1$ $ms^-1$ at a height of $2.5 xx 10^2$ $m$ . How far does it travel before it hits the ground?

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A ball is projected with a horizontal velocity of $5.1$ $ms^-1$ at a height of $2.5 xx 10^2$ $m$ . How far does it travel before it hits the ground?

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Answer 1 · 2017-06-24T13:10:37

The horizontal distance traveled will be $s=u_xt=5.1xx7.1=36.2$ $m$ .

Explanation:

As for all projectile motion questions, we can treat the horizontal and vertical direction independently. If we ignore air resistance, the horizontal velocity will remain constant at $u_x=1.5$ $ms^-1$ .

We need to calculate the time taken for the ball to fall. The initial velocity $u_y$ in the vertical direction is $0$ $ms^-1$ .

$s = u_yt +1/2a_yt^2$

$t = sqrt((2s)/a)$ (since $u_y=0$ and we can rearrange to make $t$ the subject)

$t = sqrt((2(2.5xx10^2))/9.8)=7.1$ $s$

The horizontal distance traveled in this time will be $s=u_xt=5.1xx7.1=36.2$ $m$ .

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A ball is projected with a horizontal velocity of $5.1$ $ms^-1$ at a height of $2.5 xx 10^2$ $m$ . How far does it travel before it hits the ground?

1 Answer

Explanation:

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