Solving Projectile Problems: Time, Height, and Speed Calculations

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SUMMARY

The discussion focuses on solving projectile motion problems involving a sand bag thrown by Eddy. The bag is released from a height of 1.5 m with an initial speed of 7.5 m/s at an angle of 80 degrees, targeting a vertical post 4 m high and 1.3 m away. Key calculations include the time of flight, maximum height reached, and final speed before landing. The time of flight is determined to be 1.0 s, the maximum height reached is approximately 4.3 m, and the final speed just before landing requires further calculation using vector addition of horizontal and vertical velocities.

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  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of vector addition
  • Basic algebra skills for solving equations
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This discussion is beneficial for students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to enhance their teaching methods in these topics.

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Homework Statement


Eddy throws a little sand bag so that it lands on the top of a vertical post that is 4 m high. the post is 1.3 m away from Eddy. He releases the bag from a height of 1.5 m above the ground, as shown in the figure. the initial speed of the bag is v = 7.5 m/s, the angle, theta, between the velocity and the horizontal is, theta = 80 degrees. you can neglect the friction due to the air.

1) how long does the sand bag stay in the air?

(a) t = 0.7 s
(b) t = 1.0 s
(c) t = 2.5 s

2) During its course in the air, the sand bag reaches a maximum height H. Calculate H

(a) H = 4.3
(b) H = 4.7
(c) H = 5.2

3) What is the speed of the bag just before it lands on the top of the post?

(a) 0 m/s
(b) 0.3 m/s
(c) 1.4 m/s
(d) 1.9 m/s
(e) 2.7 m/s

Homework Equations


Vf ^2 = Vo^2 - 2gH
V = /\x/t


The Attempt at a Solution



i answered #1 and #2

1) t = 1.0 s because horizontal velocity = Vox = Vx = 7.5cos80 = 1.3 m/s
/\ x = 1.3 m so
t = /\x / Vox so
t = 1.3/1.3 = 1 s

2) when velocity is 0 its at its highest point so
Vfy ^2 = Voy^2 - 2gH
*some algebra*
H = (Voy^2) / (2*9.8) = 2.78 m and its thrown at 1.5 initially so 2.78+1.5 = about 4.3 m high :)

3) IDK! :( HELP
 
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You know part of it already. You know horizontal velocity.

So figure the vertical velocity. (How fast will it be going in dropping from the max height to the landing zone on the post?)

Now just add them like vectors using the Pythagorean relationship.
 

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