They were going at a velocity 4.07m/s
<u>Explanation:</u>
Distance s =5 m
initial velocity u= 0.8 m/s
Acceleration a =1.6m/s2
We have to calculate the velocity with which they were going afterwards i.e final velocity.
Use the equation of motion
They were going with a velocity 4.07 m/s afterwards.
Answer:
-v/2
Explanation:
Given that:
<u>We know, kinetic energy is given as:</u>
consider this to be the initial kinetic energy of the body.
<u>Now after collision:</u>
Considering that the mass of the body remains constant before and after collision.
Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.
Hi there!
We can begin by calculating the time the ball takes to reach the highest point of its trajectory, which can be found using the following:
Where:
tmax = (? sec)
vsinθ = vertical comp. of velocity = 10sin(48) = 7.43 m/s)
g = acceleration due to gravity (9.8 m/s²)
We can solve for this time:
When the ball is at the TOP of its trajectory, its VERTICAL velocity is equivalent to 0 m/s. Thus, we can consider this a free-fall situation.
We must begin by solving for the maximum height reached by the ball using the equation:
d = displacement (m)
vi = initial velocity (7.43 m/s)
a = acceleration due to gravity
d = displacement (m)
y0 = initial VERTICAL displacement (28m)
Plug in the values:
Now, we can use the rearranged kinematic equation:
Add the two times together: