Question

A stone is thrown vertically upward with a velocity of 40m/second and is caught back. takingg=10m/sec2. calculate the maximum height reached by the stone. what is the net displacement and total distance covered by the stone

Answer 1

For vertical motion maximum height = u²/2g

=40²/(2×10)

=80 m

Net displacement =0 as it comes back from where it started

total distance = 2×maximum height

= 2×80

=160 m

Answer 2

where V = 0

U = 40 m/s

a = -g m/s² = -10 m/s²

0 = (40)² -2 × 10 × S

• S = 80 m .