Projectile Motion: Firing range

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1. Homework Statement

The problem and all known information is attached.

Homework Equations


The kinematic equations are attached.

The Attempt at a Solution


My attempt is attached. I did part (a), (b), and (c), right.

How do I do part (d)? I tried distance/ time .
 
Attachments
  • Handwritten physics solution on lined paper: projectile velocity components vx = 36.6 cos 42.2° and vy = 36.6 sin 42.2°, a vertical y-position equation with vertex time t = 2.5 s, and a sketch of a parabolic path with parts a) to d).
    image.jpg
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  • Photo of a printed textbook problem 27 (II): a projectile fired at 36.6 m/s at 42.2 degrees above horizontal on a flat firing range, asking for maximum height, total time in air, range, and speed 1.50 s after firing.
    image.jpg
    33.6 KB · Views: 530
  • Textbook page titled 3-6 Solving Projectile Motion Problems, showing Table 3-1 of general kinematic equations for constant acceleration in x and y components, and Table 3-2 of projectile motion equations with a_y = -g = -9.80 m/s^2.
    image.jpg
    90.4 KB · Views: 539
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Maybe you can find its vertical and horizontal velocity at that time, then you can reach the goal. Since you have done the first three problems, this couldn't be complicated for you. Give it a try!
 
Thanks for reply. Its the easiest question but I don't see it. I tried a few a things before you commented.

Can you explain why is it 28.9 m/s?
 
Last edited:
I will help. Just give me a sec to read the question.
 
You are given the initial vertical and horizontal velocity ##v_v=v_0\sin\theta## and ##v_h=v_0\cos\theta## and the vertical acceleration ##a=-g.## Because ##v_h## remains constant, so when ##t=1.5,## ##v_h=v_0\cos\theta## and ##v_v=v_0\sin\theta+at=v_0\sin\theta-1.5s\cdot g.## This can give you any idea?