If a ball is thrown vertically upward with initial velocity of 8 meters per second, what would the speed be when it returns to the starting point with acceleration of 9.81?
it returns with a velocity of 8 meters/sec. It returns with the same total energy it left with.
To answer this question, we can apply the principle of conservation of energy. When the ball reaches its highest point, all of its initial kinetic energy will be converted into potential energy. Then, as the ball falls back down, it will regain its initial kinetic energy.
Initially, the ball is thrown vertically upward with an initial velocity of 8 meters per second. We need to find the height it reaches using the equation for the maximum height of a projectile:
h = (v^2) / (2g)
Where:
h is the maximum height
v is the initial vertical velocity
g is the acceleration due to gravity
Plugging in the given values, we get:
h = (8^2) / (2 * 9.81)
h = 0.654 meters
Now that we know the maximum height, we can calculate the final speed when the ball returns to the starting point. Since the ball falls back down, it undergoes free fall. The equation for the final speed of an object in free fall is:
v^2 = u^2 + 2gh
Where:
v is the final velocity (which we need to find)
u is the initial velocity (which is 8 m/s in the opposite direction)
g is the acceleration due to gravity (9.81 m/s^2)
h is the maximum height (0.654 m)
Plugging in the values, we get:
v^2 = (8^2) + 2 * 9.81 * 0.654
v^2 = 64 + 12.86
v^2 = 76.86
v = √76.86
v ≈ 8.78 meters per second
Therefore, when the ball returns to the starting point, its speed will be approximately 8.78 meters per second.