Given :
Initial velocity , u = 0 m/s .
Acceleration due to gravity on moon , .
Height , h = 2 m .
To Find :
Final position after falling for 1.5 seconds .
Solution :
We know , by equation of motion :
Here , .
So , equation will transform by :
Therefore , the height form moon's surface is 1.88 m .
Hence , this is the required solution .
Answer:
The average acceleration is 16.6 m/s² ⇒ 1st answer
Explanation:
A rocket achieves a lift-off velocity of 500.0 m/s from rest in
30.0 seconds
The given is:
→ The initial velocity = 0
→ The final velocity = 500 meters per seconds
→ The time is 30 seconds
Acceleration is the rate of change of velocity of the rocket
→
where a is the acceleration, v is the final velocity, u is the initial velocity
and t is the time
→ u = 0 , v = 500 m/s , t = 30 s
Substitute these values in the rule
→ m/s²
<em>The average acceleration is 16.6 m/s²</em>
Answer:
In 0.5 seconds.
Explanation:
The time would be the same because it only depends on the height and the vertical component of the initial velocity. This is of course because each direction must be treated independently. Since between both cases only the horizontal speed changes, the height is the same and the vertical component of the initial velocity is null for both, the time to fall is the same.