Answer:
The second snowball hits the ground with a kinetic energy of 100 Joules
Explanation:
Given that,
From the edge of a roof you throw a snowball downward that strikes the ground with 100 J of kinetic energy. It is a case of conservation of energy.
At the highest point,
At lowest point,
From above two equation, we get :
Kinetic energy, K = 100 J
So, the second snowball hits the ground with a kinetic energy of 100 Joules. So, the correct option is (A).
Complete question is;
Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.
Block 1 is removed without disturbing block 2.
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.
Answer:
a = (m2)g/(m3 + m2)
Explanation:
Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;
T = (m3)a - - - (eq 1)
where;
T is the tension in the string
a is acceleration
m3 is mass of block 3
Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a
Making T the subject gives us;
T = (m2)g - (m2)a - - - (eq 2)
where;
g is acceleration due to gravity
T is the tension in the string
a is acceleration
m2 is mass of block 2
To solve for the acceleration, we will just substitute (m3)a for T in eq 2.
Thus;
(m3)a = (m2)g - (m2)a
(m3)a + (m2)a = (m2)g
a(m3 + m2) = (m2)g
a = (m2)g/(m3 + m2)
Answer:
The initial speed of the soccer ball is 16.38 m/s
Explanation:
given;
vertical distance y = 2.44 m
horizontal distance x = 10.0 m
angle of projection θ = 25.0°
Initial velocity has two components, Vₓ and V
Vₓ = V cosθ
V = V
sinθ
The horizontal distance = x = Vₓt + ¹/₂ ˣ g ˣ t², but g =0
x = Vₓt = V cosθ *t
10 = V cos25 *t
10 = 0.906V*t
V*t = 10/0.906 = 11.038 m
The vertical distance (g = - g, because it upward motion against gravity)
y = V*t -¹/₂ ˣ g ˣ t²
2.44 = (V sinθ)t - ¹/₂ ˣ 9.8 ˣ t²
2.44 = (V*t)sinθ - ¹/₂ ˣ 9.8 ˣ t²
2.44 = (11.038)sin25° - 4.9t²
2.44 = (11.038)*0.4226 - 4.9t²
2.44 = 4.6647 - 4.9t²
4.9t² = 4.6647 - 2.44 = 2.2247
t² = 2.2247/4.9
t² = 0.454
t = √0.454
t = 0.674 s
Recall that V*t = 11.038 m
V*0.674 = 11.038 m, solve for V
V = 11.038/0.674
V = 16.38 m/s
Therefore, the initial speed of the soccer ball is 16.38 m/s