Question

A stone is thrown vertically upward with a speed of
40 m/s. The time interval for which particle was
above 40 m from the ground is (g = 10 m/s2)
(1) 4 root 2s
(2) 8 s
(3) 4 s
(4) 2 root 2s

pls help with steps

Answer 1

Explanation:

• d= vo+1/2at2
• 40=40+1/2t2
• 40/40=10/2t2
• 1=5t2
• 1/5=t2
• 0.2=t2
• t=0.4s

Answer 2

Answer:

The stone moves by uniformly accelerated motion, and its vertical position at time t is described by the following law :]

$=> \sf y(t) = h + v_{0}t - \dfrac{1}{2} gt^2$

$\tt {\pink{Where}}\begin{cases} \sf{\green{h= 40 \: m \: is \: the \: initial \: height \: from \: which \: the \: stone \: starts \: its \: motion}}\\ \sf{\blue{v_{0}= 50 \: m/s \: is \: the \: initial \: velocity \: of \: the \: stone}}\\ \sf{\orange{g= 10 \: m/s^2 \: is the \: gravitational \: acceleration}}\end{cases}$

The time in which the stone reaches the ground is the time t at which the vertical position y(t) becomes zero:

$=> \sf 0 = h + v_{0}t - \dfrac{1}{2} gt^2$

$=> \sf 0 = 40 + 40t - 5t^2$

$=> \sf 5 {t}^{2} - 40t + 40= 0$

Dividing all the values by 5 we get :

$=> \sf {t}^{2} - 8t + 8 = 0$

By Using quadratic formula we get :

$=> \sf t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$=> \sf t = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(8)}}{2(1)}$

$=> \sf t = \dfrac{8 \pm \sqrt{64-32}}{2}$

$=> \sf t = \dfrac{8 \pm 4 \sqrt{ 2}}{2}$

$=> \sf t = 4 \pm 2 \sqrt{2}$

Therefore,, we can say that the stone reaches the ground after t = 4 ± 2 √2 seconds.