When a golfer hits a ball from the tee, the ball is in contact with the club for $0.0005 s$ . If the ball leaves the tee at a speed of $70 m / s$ , what was the acceleration of the ball?

700000 m / s^{2}

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140000 m / s^{2}

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150000 m / s^{2}

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500000 m / s^{2}

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Solution

The correct option is B $140000 m / s^{2}$
$u = 0 m / s$
$v = 70 m / s$
$a = ?$
$t = 0.0005 s$

As we know, $v = u + a t$
or, $a = \frac{v - u}{t}$
or, $a = \frac{70 - 0}{0.0005}$
or, $a = \frac{70 \times 10000}{5} m / s^{2}$
$= 140000 m / s^{2}$

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When a golfer hits a ball from the tee, the ball is in contact with the club for 0.0005s. If the ball leaves the tee at a speed of 70m/s, what was the acceleration of the ball?

The correct option is B 140000m/s2u=0m/s v=70m/s a=? t=0.0005s As we know, v=u+at or, a=v−ut or, a=70−00.0005 or, a=70×100005m/s2 =140000m/s2

When a golfer hits a ball from the tee, the ball is in contact with the club for $0.0005 s$ . If the ball leaves the tee at a speed of $70 m / s$ , what was the acceleration of the ball?

The correct option is B $140000 m / s^{2}$
$u = 0 m / s$
$v = 70 m / s$
$a = ?$
$t = 0.0005 s$

As we know, $v = u + a t$
or, $a = \frac{v - u}{t}$
or, $a = \frac{70 - 0}{0.0005}$
or, $a = \frac{70 \times 10000}{5} m / s^{2}$
$= 140000 m / s^{2}$