A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):
v^2 = u^2 + 2as
v^2 = 17.0^2 -2(9.81)(11.0)
v = √73.18 = 8.55m/s
now from equation (3):
v = u + at
8.55 = 17.0 – 9.81t
t = (8.55 – 17.0)/(-9.81) = 0.86s
To solve this problem we will apply the concepts related to the double slit-experiment. Under this concept we understand the relationship between the minimum angle, depending on the order of the fringes, the wavelength and the distance between slits. Therefore we have the following relation,
Here,
m = Order of the fringes
D = Distance between slits
= Wavelength
Replacing with our values we have,
Through the relationship between distances then we have that the basic amplification distance is given by the relationship between the distance of the slit L and the angle, then
Thus the width of the central maximum is
Therefore the widht is 0.466m
Line drive would be more farthest from the very high angle shot because when we increase the angle of flight the range started to increase and at some point it becomes maximum and that angle is 45° after that as you go on increasing the angle it won't be covering more distance as compared to the max at (45°) .
Volume is length times width times height, so 6 times 9 times 5 is 270 centimeters.
