how do you solve for the highest point in projectile motion?

Question

"A 1.5 N ball is thrown at an angle of 30o above the horizontal with an initial speed of 12 m/s. At its highest point, the net force on the ball is:
a. 14.7 N down
b. 9.8 N, 30 degrees below the horizontal
c. zero
d. 9.8 N up
e. 1.5 N down
is there equation for the highest point? I thought it is zero because it has only constant horizintal velocity, so acceleration is zero and net force is zero. I got wrong. Anybody please explain to me?

Answer 1

e. 1.5 N down.

This force is present before, during, and after the projectile is in flight. Likewise, acceleration of gravity is acting anytime. So, Simon is right.

Presumably, there are other forces acting on the projectile as it is being shot. But gravitation simply can't just be avoided. Also, when the projectile hits ground (or its intended target), gravity may be counteracted by some force, or forces, as to render it motionless. But gravitation is present, nevertheless.

Your reasoning is correct, as far as you consider horizontal motion only: horizontal velocity is constant, so horizontal acceleration is nil. Therefore, no net (horizontal) force acts on projectile. But, in your analysis, you're ignoring vertical forces altogether.

Finally, yes, there's a way to compute projectile coordinates at the instant the highest point is reached. At this point, vertical velocity ("final" velocity) is zero (otherwise, it wouldn't be the highest point; think about it). Now, by definition, v − vo = at = -gt, where v = 0 at highest point, and vo is to be the vertical component of initial velocity, i.e., vo sin θ. Time elapsed, then, can be found solving for t in -vo sin θ = -gt, or t = (vo/g) sin θ.

The x-coordinate is then x = vo cos θ · (vo/g) sin θ = (vo²/g) cos θ sin θ = (vo²/2g) sin 2θ.

Likewise,
y = vo sin θ · (vo/g) sin θ − ½ g [(vo/g) sin θ]²
y = (vo²/g) sin θ − ½ g (vo²/g) sin θ = (vo²/2g) sin² θ.

In this case, vo = 12 m/s, θ = 30°, so
x = 6.36 m; y = 1.837 m.

Answer 2

e. 1.5 N down , all the time except at t=0


...
update:
The acceleration is always 9.8 m/s^2 except at t=0.
the force at the highest point is zero, ( the acceleration is 9.8 m/s^2 )

Answer 3

1.5N down, gravity is ALWAYS pulling down