- #1
fisselt
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Last problem I can't get around..
A ball is thrown upwards at some velocity. 2 seconds later the ball is caught 10m up on a building. What is the highest possible height of the ball?
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where X_f=10)
f(t)=v_x(t)-(9.8m/s^2)t^2 (set to 0)
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where t=1.5)
I have X_f set to 10 and find the velocity of 14.8m/s. This means that the maximum height was 11.175m. This seems a little off as the ball should have dropped 0.05meters farther than the 1.175meters from my calculations. Am I making a mistake with the first equation?
Homework Statement
A ball is thrown upwards at some velocity. 2 seconds later the ball is caught 10m up on a building. What is the highest possible height of the ball?
Homework Equations
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where X_f=10)
f(t)=v_x(t)-(9.8m/s^2)t^2 (set to 0)
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where t=1.5)
The Attempt at a Solution
I have X_f set to 10 and find the velocity of 14.8m/s. This means that the maximum height was 11.175m. This seems a little off as the ball should have dropped 0.05meters farther than the 1.175meters from my calculations. Am I making a mistake with the first equation?