Question

11) An object is released from a height.
a) Find its speed at (1) t=1sec , (2) t=2sec , (3) t= 3sec
b) Find the distance travelled at 1) t = 1 sec,(2) t=2sec , (3) t=3 sec​

Answer 1

Answer:

object is released which means it is a free fall.

so 9.8 m/s

(gravity will accelerate a falling object so its velocity increases 9.8 m/s)

speed

at 1 sec. 9.8 × 1 = 9.8 m/s

2 sec. 9.8 × 2 = 19.6 m/s

3 sec. 9.8 × 3 = 29.4 m/s

distance

1 sec 9.8 m

2 sec 19.6 m

3 sec 29.4 m

Answer 2

Answer : I first tell answer which is

1. v=10 m/s 2. V=20 m/s 3. V=30 m/s

1. S=5 m 2. S= 20m 3. S= 45 m

Explanation.........

U = 0 m/s (in these type of question where object thrown from up to down u=0)

V= u + gt

V= 0 + 10(1) ------------1

V = 10 m/s

V= u +gt

V= 0 + 10 (2)

V = 20m/s

V= 0 + 10(3)

V= 30m/s

V2-U2 = 2gh

(10)2 = 2 × 10 × h

100 ÷ 20 = h

5 m = h

Like this method do for all times

400÷20 = h

20m = h

900 ÷ 20 = h

45 m =h

Mark it as brainliest...................

Thanks heart dedena yaroo...