Question #68f57

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Question #68f57

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Answer 1 · 2017-09-28T10:55:49

1) Velocity after 2 second
$v = 10.4 \frac{m}{s}$ $v=10.4 m/s$

2) Maximum Height Attained
$s = 45.91 m$ $s=45.91 m$

3) Time Taken to reach the Maximum height
$t = 3.06 sec$ $t=3.06 sec$

4) Time taken to the ball to return to the ground
$T = 6.06 sec$ $T=6.06 sec$

Explanation:

Initial Velocity $u = 30 \frac{m}{s}$ $u=30 m/s$
For upward motion Acceleration due to gravity $a = - g = - 9.8 \frac{m}{s^{2}}$ $a=-g=-9.8 m/s^2$
1) Velocity after 2 second
$t = 2 sec$ $t=2 sec$
Using Newton's First Equation of Motion
$v = u + a t$ $v=u+at$
$v$ $v$ is the final velocity after t seconds.
$v = 30 - g t = 30 - (9.8) (2) = 30 - 19.6 = 10.4 \frac{m}{s}$ $v=30-g t=30-(9.8)(2)=30-19.6=10.4 m/s$
$v = 10.4 \frac{m}{s}$ $v=10.4 m/s$

2) Maximum Height Attained
On the Maximum height final velocity of object will be zero.
$v = 0$ $v=0$
Using Newton's Third Equation of Motion
$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$
s is distance traveled
$\Rightarrow 0 = 30^{2} - 2 (9.8) s$ $=>0=30^2-2(9.8)s$
$19.6 s = 900$ $19.6s=900$
$s = \frac{900}{19.6} = 45.91 m$ $s=900/19.6=45.91 m$

3) Time Taken to reach the Maximum height
On the maximum height final Velocity $v = 0$ $v=0$
Using Newton's First Equation of Motion
$v = u + a t$ $v=u+at$
$v$ $v$ is the final velocity after t seconds.
$0 = 30 - g t = 30 - (9.8) (t)$ $0=30-g t=30-(9.8)(t)$
$9.8 t = 30$ $9.8t=30$
$t = \frac{30}{9.8}$ $t=30/9.8$
$t = 3.06 sec$ $t=3.06 sec$

4) Time taken to the ball to return to the ground
The time taken to reach maximum height and return to the ground is the same.
Hence Total time $= 2 \times$ $=2xx$ time taken to reach the maximum height
$T = 2 \times 3.06 = 6.06 sec$ $T=2xx3.06=6.06 sec$

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Question #68f57

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