Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2
h=[12×sin 610×0.473]+[−9.8×(0.473)2]
u can simplify this and u will get the answer
h=.5Gt2
H=1.09m
Answer:
<em> B.0</em>
Explanation:
Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s
Change in momentum = M(V-U)......................... Equation 1
where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.
Let: M = m kg and V = U = v m/s
Substituting these values into equation 1
Change in momentum = m(v-v)
Change in momentum = m(0)
Change in momentum = 0 kg.m/s
<em>Therefore the momentum of the ball has not changed.</em>
<em>The right option is B.0</em>
Missing question: "What is the spring's constant?"
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:

When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is

And by using Hook's law, we can find the constant of the spring:
A. Moving with constant non-zero speed
The movements of the tectonic plates