Hello!
The answer is:
The time took for the rock to reach its maximum height is 0.110 seconds.
Why?
In order to calculate the time needed for the rock to reach its maximum height, we need to calculate the initial vertical speed.
From the statement we know that the rock was launched at an initial speed of 2.1 m/s at an angle of 30° above the horizontal, so, calculating we have:
[tex]V_x=2.1\frac{m}{s} *cos(30\°)=1.82\frac{m}{s} \\\\V_y=2.1\frac{m}{s} *sin(30\°)=1.05\frac{m}{s}[/tex]
Also, we know that at the maximum height, the speeds tends to 0. So, using the following equation and substituting "v" equal to 0, we have:
[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]
Where,
t is the time in seconds.
V is the initial speed
g is the gravity acceleration.
Using gravity acceleration equal to [tex]9.81\frac{m}{s^{2} }[/tex] we have:
[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]
[tex]t=\frac{V_o}{g}[/tex]
[tex]t=\frac{1.05\frac{m}{s}}{9.81\frac{m}{s^{2}}}=0.1070seconds=0.110seconds[/tex]
Hence, the correct option is the last option, the time took for the rock to reach its maximum height is 0.110 seconds.
Have a nice day!
