Question

A stone is thrown upward from the top of a 59.4m high cliff with an upward velocity component 19.6 m/s .How long is stone in the air?

Answer 1

Answer:

taking equation

Vf=Vi+at

0=19.6-9.8t

T1=19.6/9.8=2s

now taking equation

S1=Vo+1/2at^2

S1=19.6(2)-1/2(9.8)(2^2)

S1=19.6m

Total height reached after 2sec is

H=h1+s1

H=59.4+19.6=79m

T2=time to fall through 79m

so, h=Vit+1/2at^2

79=0+1/2×9.8×t^2

79/4.9=t2^2

t2^2=790/49

t^2√16

t2=4s

Total time is =t1+t2=6sec

Answer 2

The stone was for 6sec in the air.

Explanation:

Applying newtons formula,

S= ut + (1/2)at^2

here a=9.8m/s²

Let downwards direction be (-)

then,

-59.4 = 19.6t – (1/2) 9.8t²

-59.4 = 19.6t - 9.8/2 t²

-9.8/2 t² + 19.6 t + 59.4 = 0

solving for t we get,

x =  -b ± √b²−4ac / 2a

(-19.6 ± √ (19.6)² - 4 * (-9.81)/2 * 59.4 ) / 2 * (-9.8)/2

-19.6/-9.8 ± (√384.16 + 1165.428) / (-9.8)

2 ±  4.008

case1 : 2 + 4.008

        = 6.00

case 2:  2 - 4.008

          = -2.00 (neglected because the time period can't be negative)

t= 6 sec

Time is 6 sec.

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