a)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = final position of stone = 20.0 meters
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = ?
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
v² = 30² + 2 (- 9.8) (20 - 0)
v = 22.5 m/s
b)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = maximum height gained
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = 0 m/s
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
0² = 30² + 2 (- 9.8) (Y - 0)
Y = 46 m
<span>Anger is to angry as fire is to blazing. </span>
I think it’s C. Stroke if not then D
Answer:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²