Another physics problem!!?

Question

A projectile is fired from a gun and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s respectively.

Approximately how long does it take the projectile to reach the highest point in its trajectory?

1s

2s

4s

8s

16s

What is the speed of the projectile when it is at the highest point in its trajectory?

0 m/s

20 m/s

30 m/s

40 m/s

50 m/s

Answer 1

this is easy, do it urself.  start by writing down what was given and what is assumed, like the acceleration is -9.81 m/s2.  i usually make a T-chart with the vertical and horizontal components.

Answer 2

t = v/g = 40/9.8 ~= 4 s.

At the highest point, vertical velocity is zero.

=> the projectile has only the horizontal velocity

=> its speed will be 30 m/s.

Answer 3

I'm afraid my brain is too much of mush right now to answer the first, question, the but second, I believe, should be fairly obvious following what I believe to be correct logic assuming, of course, that air resistance is negligible: at the highest point in the trajectory, the projectile will have absolutely no vertical speed, but it will, however retain its horizontal speed of 30m/s, so that should be your correct answer there.

Hopefully somebody with a more active brain can chime in for the first question.