The maximum height of the lollipops can be calculated using the kinematic equation ( \text{max height} = \frac{v^2}{2g} ), where ( v = 3ms ) and ( g = 9.81ms^2 ) (acceleration due to gravity). Plugging in the values, we get ( \text{max height} = \frac{3^2}{2 \times 9.81} \approx 0.46m ).
The ball has the highest gravitational potential energy when it is at its highest point in the air, as that is when it has a velocity of zero and is up the highest.
Multiply the height by 4. The equation to use is h=-16t2 + v0t + h0. Use whatever values you want for v0 and h0, and find the vertex of the parabola. Then double your value of v0, and find the vertex of your new parabola. It will be 4 times as high every time. By the way, to find the vertex, plug in v0/32 for t. Then solve for h.
To have zero speed at the top, you need to throw the projectile with an initial velocity such that it reaches its maximum height at that point. This requires the initial velocity to be exactly equal to the velocity that would be attained due to gravity when the projectile falls from that height. The angle of projection should be such that the vertical component of the initial velocity cancels out the velocity due to gravity.
Anyone who is trying to figure out how high they can vertically throw an object like a ball has a little math to do. The calculation needed is the time it takes the ball to reach its max height multiplied by the rate at which gravity kicks in. Take that number and divide it in half.
At the very top of the ball's path, its velocity is momentarily zero. This is because as the ball reaches its maximum height, it stops moving vertically before falling back down due to gravity.
The ball has the highest gravitational potential energy when it is at its highest point in the air, as that is when it has a velocity of zero and is up the highest.
Multiply the height by 4. The equation to use is h=-16t2 + v0t + h0. Use whatever values you want for v0 and h0, and find the vertex of the parabola. Then double your value of v0, and find the vertex of your new parabola. It will be 4 times as high every time. By the way, to find the vertex, plug in v0/32 for t. Then solve for h.
To have zero speed at the top, you need to throw the projectile with an initial velocity such that it reaches its maximum height at that point. This requires the initial velocity to be exactly equal to the velocity that would be attained due to gravity when the projectile falls from that height. The angle of projection should be such that the vertical component of the initial velocity cancels out the velocity due to gravity.
This problem can be solved simply by using the following equation: Vf^2= Vi^2+2*a*d Vf = 0 at the maximum height Vi = 95 ft/s a = acceleration due to gravity = -32 ft/s^2 Plugging these in gives a maximum height from the beginning of the throw as: d= 141 ft However the ball started from a height of 145 ft so the ball reaches a maximum ht of 286 ft.
Anyone who is trying to figure out how high they can vertically throw an object like a ball has a little math to do. The calculation needed is the time it takes the ball to reach its max height multiplied by the rate at which gravity kicks in. Take that number and divide it in half.
At the very top of the ball's path, its velocity is momentarily zero. This is because as the ball reaches its maximum height, it stops moving vertically before falling back down due to gravity.
As velocity never exceeds the velocity of light.... so i hope a man running with the velocity of light will not be able to throw a ball with any velocity.......... we may get the maximum n minimum velocity with which that can be thrown mathematically that we may get it to be zero................
v2=sqrt(2*g*h) v...starting velocity g...gravity h...height achieved
The boy throws a stone vertically upward with an initial velocity of 6.0 m/s, meaning the stone is moving against gravity. It will reach a maximum height and then fall back down due to gravity. The stone will eventually return to the ground after reaching its highest point.
You throw a ball straight up with a velocity of 40 meters per second. What is the ball's velocity after 3 seconds?
To use a velocity ball, hold it in your hand and throw it with as much force as possible. The ball is designed to measure the speed of your throw. Make sure to take into account your technique and form when using the velocity ball to get an accurate reading of your throwing speed.
Yes, but only for an instant. Throw a rock or a baseball straight up. After a while, it reaches its maximum height, stops moving up, and starts moving down. From the time it leaves your hand, its acceleration is constant. But at the very top of the arc, where it's changing from moving up to moving down, at that instant, it has zero velocity.