questions about falling fragments from fireworks?

Question

If you shoot a firework into the air and it hits it's maximum height in 7 seconds and explodes sending fragments scattering in horizontal directions all around at a speed of 20 meters per second then (assume that air drag is negligible):

It would only take the fragments 7 seconds to fall right?
A)What will be the acceleration of the fragments as they fall vertically?

B)What is the path of the fragments as they fall to the ground?

C)Why do the fragments follow this path?

Please explain how you figured these out.I'm trying to learn but it's so confusing.

Answer 1

Air drag would not be negligible. However if it was, It wouldn't necessarily take 7 seconds to fall. This would only be the case if the firework accelerated up constantly at 9.81m/s2. It could be more or less than this.
The acceleration vertically (downwards) would be 9.81m/s2 like everything else that falls
the path would be a parabola effectively it is a y=x^2 as the horizontal velocity is linear (proportional to t) and the vertical velocity is a function of t^2

Again air drag has a huge effect. Acceleration down would stop in about 0.5 second.

Answer 2

Amania’s answer is helpful, but let me give you additional data.
You ask: “It would only take the fragments 7 seconds to fall right?” – YES
No matter how they are falling their acceleration is g coz there is no other force but gravity.
Thus the horizontal path X=v*t, v=20m/s, t=7s, X=140m.
The vertical path Y=g*t^2/2=240.1m, the whole path along parabola S > than sqrt(X^2+Y^2).
But I think you are not yet taught integration.