Question

From the top of a 490 m high cift, a boy throws a stone horizontally with an initial speed
15 ms. What is the time taken by the stone to reach the ground?
In the above problem
find the speed with which the stone hits the ground​

Answer 1

Answer:

t=8.34 s and v= 96.732m/s

Explanation:

To find the time we use the 2nd equation of motion $s= ut + 1/2gt^2$

Here u= 15 m/s , s=490 m and g=9.8 m/s^2

$490 = 15t +1/2*9.8t^2\\490= 15t +4.9t^2$

$4.9t^2 + 15t - 490 = 0$

Applying t= -b +/- √(b^2 -4ac)/ 2a

$t=-15 +/-$ √$15^2- 4*4.9*490$$/2*4.9$

$t = -15 +/-$√$225-9604$/2*4.9

$t =-15 +/-$ √$9379$$/ 9.8$

$t= -15+96.8/9.8$

t= 81.8/9.8

t=8.34 s

Now speed,

$v=u+gt$

$v= 15+9.8*8.34$

$v=15+ 81.732 =96.732m/s$