Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
Answer:
Weight
Explanation:
The spring balance is used to measure weight of an object.
Compare the initial mass to the final mass.
Answer:
200 N
Explanation:
Given that,
A ball traveling at 15 m/s hits a bat with a force of 200 N.
We need to find the force that the bat moving at 20 m/s hit the ball with.
We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.
It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).
