Question

a stone is thrown vertically upward with a speed of 5 metre per second how high does the stone rise before returning back to the earth give acceleration due to Earth attraction G is equal to 9.8 metre per second square ​

Answer 1

Answer:

1.28m

Explanation:

We know that at the highest point final velocity is 0

∴ We can find out the time needed to reach the highest point by the following equation-

$v=u+at$ , the following equation can be modified as -

$v=u-gt$ (∵ $g$ is always directed downwards and we have taken the downwards direction as negative)

⇒$0=5-9.8 \cdot t$

⇒$-5=-9.8 \cdot t$

⇒$t = \frac{5}{9.8} = 0.51$

Now we can substitute the value of $t$ in the second equation of motion $s=ut-\frac{1}{2} gt^{2}$

⇒$s=5 \cdot  0.51 -\frac{1}{2}  \cdot 9.8 \cdot 0.51^{2} = 2.55 - 1.27 =1.28$

∴ $Ans = 1.28m$

(Please ignore the A)

Answer 2

Answer:

hope you like it (•‿•)

Explanation:

We know that at the highest point final velocity is 0

∴ We can find out the time needed to reach the highest point by the following equation-

v=u+atv=u+at , the following equation can be modified as -

v=u-gtv=u−gt (∵ gg is always directed downwards and we have taken the downwards direction as negative)

⇒0=5-9.8 \cdot t0=5−9.8⋅t

⇒-5=-9.8 \cdot t−5=−9.8⋅t

⇒t = \frac{5}{9.8} = 0.51t=

9.8

5

=0.51

Now we can substitute the value of tt in the second equation of motion s=ut-\frac{1}{2} gt^{2}s=ut−

2

1

gt

2

⇒s=5 \cdot 0.51 -\frac{1}{2} \cdot 9.8 \cdot 0.51^{2} = 2.55 - 1.27 =1.28s=5⋅ 0.51−

2

1

⋅9.8⋅0.51

2

=2.55−1.27=1.28

∴ Ans = 1.28mAns=1.28m