Projectile motion 2-d kinematics

[jehan4141]

This is an even problem from Cutnell & Johnson 8 edition. Can anyone verify if it is correct?

In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 35.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m. Ignore air resistance.

(a) The first balloon launched by the Jackson team hits Walton dorm 2.0 s after launch, striking it halfway between the ground and the roof. Find the direction of the balloon's initial velocity. Give your answer as an angle measured above the horizontal.

(b) A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed of this second balloon.

MY WORK FOR PART A

t = 2 seconds
x = 35 meters
y = -4 meters

V_ox = x/t = 35/2
V_ox = 17.5 m/s
V_ox = V_ocosƟ
17.5 = V_ocosƟ
V_o = 17.5/cosƟ

V_oy = V_o + at

y = V_oyt + 0.5at² = -V_osinƟ -4.9t²
-4 = (17.5/cosƟ)sinƟt -4.9t² <----plug in t = 2
-4 = (17.5)(2)tanƟ - (4.9)(4)
Ɵ = 24.0 degrees

V_o = 17.5/cosƟ
V_o = 17.5/cosƟ = 19.2 m/s

ahh i forgot to do part B...But the answer is 28.9 m/s for part B?

 

[PeterO]

This is an even problem from Cutnell & Johnson 8 edition. Can anyone verify if it is correct?

In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 35.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m. Ignore air resistance.

(a) The first balloon launched by the Jackson team hits Walton dorm 2.0 s after launch, striking it halfway between the ground and the roof. Find the direction of the balloon's initial velocity. Give your answer as an angle measured above the horizontal.

(b) A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed of this second balloon.

MY WORK FOR PART A

t = 2 seconds
x = 35 meters
y = -4 meters

V_ox = x/t = 35/2
V_ox = 17.5 m/s
V_ox = V_ocosƟ
17.5 = V_ocosƟ
V_o = 17.5/cosƟ

V_oy = V_o + at

y = V_oyt + 0.5at² = -V_osinƟ -4.9t²
-4 = (17.5/cosƟ)sinƟt -4.9t² <----plug in t = 2
-4 = (17.5)(2)tanƟ - (4.9)(4)
Ɵ = 24.0 degrees

V_o = 17.5/cosƟ
V_o = 17.5/cosƟ = 19.2 m/s

ahh i forgot to do part B...But the answer is 28.9 m/s for part B?

Nice method of calculation!

 

[jehan4141]

WOOHOO! Thank you! I love this site...My teacher refuses to give us the answers for even problems, even right before the quiz! How are we supposed to know if we are doing it right :(

 

[PeterO]

WOOHOO! Thank you! I love this site...My teacher refuses to give us the answers for even problems, even right before the quiz! How are we supposed to know if we are doing it right :(

If you think you have the right answer, you are most probably right. If you are not sure whether you are right, you are quite possibly wrong. CONFIDENCE! - backed up by sound preparation.

 

[rwisz]

Yes, that is correct. To find the initial speed for part B, we can use the same equation from part A: Vo = 17.5/cosƟ. However, in this case, we are given the angle Ɵ (24.0 degrees) and the distance traveled (35.0 m), so we can rearrange the equation to solve for Vo: Vo = 35.0/cosƟ. Plugging in Ɵ = 24.0 degrees, we get Vo = 28.9 m/s.