Question

A bullet is fired horizontally with a velocity of 80 m/second square then distance during first second

Answer 1

Answer:

t =1s, v=80 m/s sq, u =0, s =?, a=9.8

By newton's 2nd law of motion,

S=ut +1/2at×t (square)

S=0×1+1/2 ×9.8 ×1

S=4.9metres

Answer 2

Given:

velocity 80 m/s

time = 1sec

To Find:

Distance during first second

Solution:

we know,

s = ut +1/2 gt²

u = 0 (initail velocity = 0 as it has just started)

s = displacement

g = 9.8 m/s²

t = 1sec

now,

s = 0*1 + $\frac{1}{2}$ * 9.8 * 1

s = 9.8 / 2 = 4.9 m

s = 4.9 m

So, distance covered during first second = 4.9m.