63 105
Assignments Done
99%
Successfully Done
In July 2018

Answer to Question #15454 in Mechanics | Relativity for Daniela

Question #15454
A watermelon cannon fires a watermelon vertically up into the air at a velocity of +12.0 m/s, starting from an initial position 1.20 meters above the ground. When the watermelon reaches the peak of its flight, what is (a) its velocity, (b) its acceleration, (c) the elapsed time, and (d) its height above the ground?
Expert's answer
(a) its velocity

The velocity of a watermelon has zero value at the peak of its flight.

(b) its acceleration

The acceleration of a watermelon has zero value at the peak of its flight too.

(c) the elapsed time

We can write down the next

V = V0 - gT,

where V is the current velocity of a watermelon, V0 = 12 m/s is the initial velocity of a watermelon, g is the gravity acceleration and T is the time elapsed after firing. As the velocity of a watermelon has zero value at the peak of its flight, we can find the time elapsed by watermelon to reach the peak:

V0 - gT = 0 ==> T = V0/g = 12[m/s]/9.8[m/s²] ≈ 1.22 s.

(d) its height above the ground?

H = H0 + gT²/2 = 1.20[m] + 9.8[m/s²]*(1.22[s])²/2 ≈ 8.49 m.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions