paulm1285 said:
danago said:9.8 m/s/s is the acceleration downwards due to the force of gravity. Where does it come into play? Well, if it didnt exist, then i don't think it would be possible to really define a maximum height as the ball would keep going upwards
You are correct about the speed being zero at the top.
sanalsprasad said:
The maximum height reached by a thrown object is affected by several factors, including the initial velocity, the angle of launch, air resistance, and gravity. The higher the initial velocity and launch angle, the higher the object will go. Air resistance will slow down the object, while gravity will bring it back down to the ground.
The maximum height reached by a thrown object can be calculated using the following formula: h = (v^2 * sin^2θ) / 2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s^2).
The optimal angle for maximum height will depend on the initial velocity of the object. However, in general, a launch angle of 45 degrees will result in the highest maximum height for a given initial velocity.
Air resistance, also known as drag, will decrease the maximum height reached by a thrown object. This is because air resistance acts in the opposite direction of the object's motion, slowing it down and reducing its maximum height. Objects with a larger surface area or irregular shapes will experience more air resistance.
No, the maximum height reached by a thrown object cannot be greater than the initial height. This is because gravity will always pull the object back down to the ground. The maximum height reached will always be equal to or less than the initial height, depending on factors such as air resistance and launch angle.