To determine the maximum height reached by the rock and the time it takes to reach that height, we can use the equations of motion for vertical motion under constant acceleration.
Given: Initial velocity (uuu) = 10 m/s (upwards) Acceleration due to gravity (aaa) = -9.8 m/s² (downwards) Height above the river (sss) = 5.0 m
To find the time (ttt) it takes to reach the maximum height, we can use the following equation:
v=u+atv = u + atv=u+at
At the maximum height, the final velocity (vvv) will be zero since the rock momentarily comes to a stop before reversing its direction. Therefore:
0=10−9.8t0 = 10 - 9.8t0=10−9.8t
Solving for ttt:
9.8t=109.8t = 109.8t=10
t=109.8t = frac{10}{9.8}t=9.810
Now we can find the maximum height (hhh) reached by the rock. We'll use the equation:
h=ut+12at2h = u t + frac{1}{2} a t^2h=ut+21at2
Substituting the values:
h=10×109.8+12(−9.8)(109.8)2h = 10 imes frac{10}{9.8} + frac{1}{2} (-9.8) left(frac{10}{9.8}
ight)^2h=10<spa