Answer:
time of collision is
t = 0.395 s
so they will collide at height of 5.63 m from ground
Explanation:
initial speed of the ball when it is dropped down is
similarly initial speed of the object which is projected by spring is given as
now relative velocity of object with respect to ball
now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m
Now the height attained by the object in the same time is given as
so they will collide at height of 5.63 m from ground
Explanation:
A particular kind of matter with uniform properties..
Answer:
1.125m/s^2
Explanation:
Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically
v^2= u^2+2as
Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.
a = ?
u = 0m/s
v = 15m/s
s = 100m
Substituting the values into the formula above
v^2= u^2+2as
15^2=0^2+2×a×100
225= 0+200a
225= 200a
Divide both sides by 200
225/200 = 200a/200
a= 1.125m/s^2
Hence the acceleration of the car is 1.125m/s^2.
Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s
I think it’s D) all of the above
