A $0.1$ kg nut is thrown by a squirrel off of the top of a $10$ m tall tree. It is thrown with an initial upward velocity of $4$ m/s and an initial horizontal velocity of $3$ m/s.
a) How long till it reaches the top of its arc?
b) From the top of its arc how long till it reaches the ground?
c) How far will it travel horizontally from the moment it is let go to when it reaches the ground?
The upward and horizontal velocity together are causing me trouble on the right formulas to use. Should I solve for both individually then add and divide the answer by 2? Can anyone recommend what formulas I should use for parts A B and C?
Also, I believe the mass of the nut is irrelevant to answer the question, is that correct?
Any help is greatly appreciated. If you could recommend the proper formulas I can do the work. Or let me know of any similar examples that you may recommend.
Thanks
Don't worry too much about there being vertical and horizontal components-- they don't really interact in this question, so it's totally fine to consider them separately, especially since the only force acting on the nut is $mg$ in the downward direction.
Notice that the horizontal velocity is completely unrelated to parts a and b. As such, what if I told you to just treat it like a 1-D problem for parts a and b, where there's no horizontal velocity and only a vertical velocity?
We can find the amount of time it takes to reach the peak of its flight by calculating at what time $t_1$ the nut has zero vertical velocity: $$v_f = 0 = v_i + at = 4.0 \frac{\text{m}}{\text{s}} + -9.81 \frac{\text{m}}{\text{s}^2}*t_1$$
Now, you likely know a formula for finding the distance traveled given an initial velocity and a constant acceleration, which looks like (unless I'm mistaken) $$d = v_i*t + \frac{1}{2}at^2 $$ So, use $t_1$ and this formula to find the upward distance traveled, and call this distance $d_1$.
With $d_1$, you'll want to calculate the time it takes the nut to fall from the top of it's peak to the ground. Don't forget that the nut started in a tree $10m$ off of the ground. We can use the second formula I mentioned, this time plugging in something like $$d_1+10 = v_i*t+\frac{1}{2}at^2 = 0*t_2 +\frac{1}{2}*(9.81)t_2^2 $$ Again, solve for $t_2$ and now you've found the amount of time it takes the nut to follow its full trajectory, being $t_{tot} = t_1+t_2$.
For part c, since the horizontal velocity is constant, we just calculate the horizontal distance traveled as $d_h = v_i*t+\frac{1}{2}at^2 = 3*t_{tot}$.