Question

a stone is thrown vertically upwards with speed 209.4m/s. find. (1) time taken by stone to reach the maximum speed (2)maximum height reached by stone

Answer 1

Answer:

solved

Explanation:

speed u =  209.4m/s , g = 10 m/s²

(1) time taken by stone to reach the maximum speed = 209.4/10 = 20.94 s

(2)maximum height reached by stone = 209.4²/20 = 2192.418 m.

Answer 2

Explanation:

here....

u= initial velocity= 209.4 m/sec

1) the speed will be maximum at the end of its journey....

so, total time taken= ?

Solution

we will first find the time of half journey when the stone is at the highest height.

we know that...

time taken to reach the max. height is equal to the time taken to return back from that point to ground.....

so, u= 209.4 m/sec

v= 0m/sec

g= -10m/sec^2

t=0

v=u+gt

0= 209.4 -10(t)

10 t = 209.4

t= 20.94 sec....for the half journey....

so, for total journey....

time taken= 20.94× 2= 41.88 sec

so, time at which the speed will be max. is 41.88 sec or 42 sec(approx)

2) max. height attained = ?

we know.......

${v}^{2} = {u}^{2} + 2gh \\ \\ h = \frac{ {v}^{2} - {u}^{2} }{2g} \\ \\ = \frac{0 - 43848}{20} \\ \\ = 2192.41 \: m$

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