It was about 9:30 p.m. sorry if the answer is wrong
<span>3.78 m
Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes.
7.2 m/s / 9.81 m/s^2 = 0.77945 s
The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving
d = 1/2 A T^2
d = 1/2 9.81 m/s^2 (0.77945 s)^2
d = 4.905 m/s^2 0.607542 s^2
d = 2.979995 m
So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height.
d = 2.979995 m + 0.8 m = 3.779995 m
Rounding to 2 decimal places gives us 3.78 m</span>
Answer:
7.6 s
Explanation:
Considering kinematics formula for final velocity as
Where v and u are final and initial velocities, a is acceleration and s is distance moved.
Making v the subject then
Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then
Also, v=u+at and making t the subject of the formula
Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then
Therefore, it needs 7.6 seconds to travel
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.
Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
The kinetic energy of an object of mass m and velocity v is given by
Let's call
the initial speed of the car, so that its initial kinetic energy is
where m is the mass of the car.
The problem says that the car speeds up until its velocity is twice the original one, so
and by using the new velocity we can calculate the final kinetic energy of the car
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.