Question

1. A stone is thrown straight up at a speed of 12 m / s.

a. After how long does the stone turn down again?

b. How high does the stone reach before it turns?

Answer 1

We have given the initial velocity $u = 12ms^{-1}$

a.) Let the time taken by the stone turn down again = t

Displacement s =0,

We know,

$s= ut+\dfrac{1}{2}gt^2$

$0 = (-12)t+ \dfrac{1}{2}(10)t^2$

$t = \dfrac{12\times2}{10}$

$t = 2.4s$

b.)Height up to the stone reach before it turns = maximum height it achieves

Max. height can be calculated using third equation of motion,

$v^2-u^2 = 2gh$

$(0)^2-(12)^2 = 2\times -10 \times h$

$h= 7.2m$

$h= H_{max} = 7.2m$