Solving Projectile Problems: Time, Height, and Speed Calculations

[Johny 5]

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

 

[LowlyPion]

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.

 

[nlightner]

I would first congratulate you on your correct answers for #1 and #2. For #3, we can use the equation Vf^2 = Vo^2 - 2gH to find the final velocity (Vf) of the sand bag just before it lands on the top of the post. We know that the initial velocity (Vo) is 7.5 m/s and the height (H) is 4 m (since it lands on the top of the post which is 4 m high). We can plug these values into the equation and solve for Vf:

Vf^2 = (7.5 m/s)^2 - 2(9.8 m/s^2)(4 m)
Vf^2 = 56.25 m^2/s^2 - 78.4 m^2/s^2
Vf^2 = -22.15 m^2/s^2
Vf = √(-22.15) m/s
Vf = 4.7 m/s (rounded to one decimal place)

Therefore, the speed of the bag just before it lands on the top of the post is 4.7 m/s. This is option (b) in the given choices.