Question

please solve this question an object thrown upward with a velocity of 40 km/ hr what is the height of the object after 20s?(g=10m/s2)

Answer 1

Given, initial velocity of the object v=40kmph
i. e
$40 \times (5 \div 18) = \frac{100}{9}$
Time t= 20sec
From
$s = ut + \frac{1}{2} a {t}^{2}$
Height of the tower = 1
$\frac{100}{9} \times 20 + \frac{1}{2} \times 10 \times 400$
11.11+2000=2011.1 m
= 2.011km

Answer 2

Given, initial velocity of the object v=40kmph  

i. e  

Time t= 20sec

From  

Height of the tower = 1

11.11+2000=2011.1 m

= 2.011km

height=2.1 km (approx)