# Acceleration question

by rmackay
Tags: acceleration
 Emeritus Sci Advisor PF Gold P: 10,429 Your formula is correct; the acceleration due to gravity is constant, so the total change in acceleration from launch to landing is gt. However, you're going about this problem the hard way, since you'd need to find t first to use your formula, and finding t in this case could be a bit complicated. How about using an energy argument instead? The ball begins with a specific amount of kinetic energy, which it trades for altitude as it rises to the top of its trajectory. It then begins to fall, and trades that altitude back into kinetic energy. Since there is no friction, no energy is lost. The ball is going the same speed when it was launched upwards from 5 m altitude as when falling downward past 5 m altitude. In other words, it's going 10 m/s downwards as it passes 5 m altitude. Now, the problem is significantly simplified! All you need to do is determine how much speed the ball would gain in falling the additional 5 meters to the ground, then add it to 10 m/s it had at 5 m altitude. This doesn't require time at all; all you need to do is figure out how much gravitational potential energy the ball lost in moving downwards 5 meters ($= mg\Delta h$). All of that gravitational potential energy is converted into kinetic energy (so $mg\Delta h = (1/2) m(\Delta v)^2$). Can you solve that equation for $\Delta v$? Hint: This is an extremely common sort of thing to be asked in a high-school physics class -- how much speed does a ball gain in falling x meters. You might want to remember the very simple formula you'll find. - Warren
 Emeritus Sci Advisor PF Gold P: 10,429 Finding the time (if you decide to go that route) involves solving the position equation: $$y(t) = y_0 + v_0 t + \frac{1}{2} g t^2$$ where $y_0 = 5$ and $v_0 = 10$. - Warren