Answer:
3.8 secs
Explanation:
Parameters given:
Acceleration due to gravity, g = 9.8
Initial velocity, u = 11.76 m/s
Final velocity, v = 49 m/s
Using one of Newton's equations of linear motion, we have that:
where t = time of flight of arrow
The sign is positive because the arrow is moving downward, in the same direction as gravitational force.
Therefore:
The arrow was in flight for 3.8 secs
Answer:
Explanation:
Yes , the horizontal distance travelled by the ball will depend upon the height of release .
When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown
depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .
Horizontal distance covered = t x horizontal velocity
If V be the velocity of throw and Vx be its horizontal component
Horizontal distance covered = t x Vx
Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .
If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity
from the relation
s = ut + 1/2 at²
h = - Vy t + 1/2 at²
As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0
h = 1/2 gt²
Horizontal distance covered = t x Vx
=
From this expression also
Horizontal distance covered is proportional to .
Answer:
Work done = -220,000 Joules.
Explanation:
<u>Given the following data;</u>
Mass = 1100kg
Initial velocity = 20m/s
To find workdone, we would calculate the kinetic energy possessed by the car.
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
Where,
Substituting into the equation, we have;
K.E = 220,000J
Therefore, the workdone to bring the car to rest would be -220,000 Joules because the braking force is working to oppose the motion of the car.