Question

A stone is thrown vertically upward with an initial
velocity of 14ms-1.find the maximum height and the time of ascent​

Answer 1

Answer:

Ht. = 10m Time of Ascent = 10/7s or 1.428s.

Explanation:

u = 14m/s

v = 0 m/s

g = 9.8 m/s²

As gravitational force is applied downwards and the ball is thrown upwards, it will be taken as negative.

Applying 3rd eq.n of Motion,

v² = u² + 2(-g)h

0² = 14² + 2(-9.8)(h)

0 = 196 +(-19.6h)

19.6h = 196

h = 196 / 19.6

h = 10m.

To find the time of ascent, apply 2nd eq.n of motion,

s = ut + 1/2 at²

10 = 14t + 1/2 (-9.8)t²

10 = 14t - 4.9t²

4.9t² - 14t + 10 = 0

4.9t² - 7t -7t +10 =0

-7t(-0.7t + 1) + 10(-0.7t +1) = 0

(-7t +10) (-0.7t + 1) =0

if -7t + 10 = 0

7t = 10

t = 10/7s

if -0.7t +1 = 0

0.7t = 1

t = 1/0.7

t = 10/7s

t= 1.428s