rejoiceotung
Answered

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3. a stone thrown up is caught by the thrower after 6 seconds how high did it go and where was it 4 seconds after start (g = 9.8m/s2)??​

Sagot :

shribhagwan568

Answer:

here, v = 0, a = - 9.8 m/sec^2, t = 3 sec

v = u + at

0 = u + (-9.8) * 3

0 = u - 29.4

u = 29.4 m/sec

here, u = 29.4m/s, t = 3sec, a= -9.8m/s^2

s = ut + 1/2 a t^2

s = 29.4 * 3 + 1/2 * - 9.8 * 3^2

s = 88.2 - 4.9 * 9

s = 88.2 - 44.1

s = 44.1 m

here, u = 0, a = 9.8m/s^2, t = 1 sec, s = ?

position after 4sec :

s = ut+ 1/2 at^2

s = 0 * 1 + 1/2 * 9.8 * 1^2

s = 4.9m

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