Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
Answer: 1200kg
Explanation:
KE = (1/2)mv^2
103kJ = 103000J
103000J = (1/2) * m * (13.1m/s)^2
Solve for m
Let's use the mirror equation to solve the problem:
where f is the focal length of the mirror,
the distance of the object from the mirror, and
the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for
by using the numbers given in the text of the problem:
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.
So, the hang up time, or the time the object stayed on air is calculated using this equation:
a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
