Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
6.5 m/s
Explanation:
We are given that
Distance, s=100 m
Initial speed, u=1.4 m/s
Acceleration, 
We have to find the final velocity at the end of the 100.0 m.
We know that
Using the formula
Hence, her final velocity at the end of the 100.0 m=6.5 m/s
If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.
When somebody hands you a Celsius°, it's easy to find the equivalent Fahrenheit°.
Fahrenheit° = (1.8 · Celsius°) + 32° .
So 100°C works out to 212°F.
It's also easy to find the equivalent Kelvin. Just add 273.15 to the Celsius.
So now you can see that 100°C is equal to A and D,
and it's less than B .
The only one it's greater than is C .
