Question 1084555: A ball is thrown straight up with an initial speed of 40m/s. How long will it take to reach the top of its trajectory and how high will the ball go?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Gravity acts to decelerate the ball at a rate of 9.81 m/s^2. What this means is that after each second, the speed has decreased by 9.81 m/s. At the top of the trajectory, the final speed is 0.
Thus the time t = (40 m/s)/(9.81 m/s^2) = 4.08 seconds.
Note that we have expressed in words what the formula v2 = v1 + a*t actually means, where in this case a = g, the acceleration due to gravity, and v1, v2 are the initial and final speeds, respectively.
To get the height, we use either the formula
h = v1*t + 1/2a*t^2 or v2^2 = v1^2 + 2a*h
Using the 2nd formula, we get h = v1^2/2/g = 40^2/2/9.81 = 81.55 m
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