Answer:
Explanation:
We shall find first the velocity of ball at the time when string breaks. Let it be v . During its fall on the ground , 1.02 m below, we use the formula
h = 1/2 gt² where t is time of fall .
1.02 = 1/2 x 9.8 x t²
t²= .2081
t = .456
During this time it travels horizontally at distance of 2.5 m with uniform velocity of v
v x .456 = 2.5
v = 5.48 m /s
centripetal acceleration
= v² / r where r is radius of the circular path
= 5.48² / .478
= 62.82 m /s²
Answer:
a = v²/r
Explanation:
The acceleration of a body moving in a circular path is known as the centripetal acceleration. This is the acceleration of a body that keeps the body within the circular path. It is written in terms of the linear velocity v and the radius of the circle of rotation as shown;
a = v²/r where
v is the linear velocity
r is the radius
a is the centripetal acceleration
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......
17.15 m/s</span>
Answer:
39.2m/s
Explanation:
The potential energy the book has right before it falls is equal to the kinetic energy in falling.
PE = KE
mgh = (1/2)mv
2gh=v
v=(2)(9.81)(2)
v=39.24m/s
The statement is false. Vectors are used to solve projectile motion problems because they allow the analysis of one direction at a time for two-dimensional motion. Scalar quantities can be used to analyze linear motion problem, but not projectile motion.
