A rock is launched from ground level at a speed of 25 m/s, 62 degrees above hori

Question

A rock is launched from ground level at a speed of 25 m/s, 62 degrees above horizontal. Let x be the horizontal axis and y be the vertical axis, with up chosen to be postive.

(a) Calculate the components of its initial velocity:
Vix=                         Viy=

(b) What are the components of its acceleration?
ax=                           ay=

(c) What is the rock's speed when it reaches its highest point?

(d) The rock lands on top of a storage tank 3.6s after launch. Calculate the height of the tank. Show all work.

Explanation / Answer

Given that

rock launhed from ground with initial velocity vi = 25 m/s , projection angle is theta = 62 degrees with the horizontal

a)

Vix = Vi cos theta = 25cos62 m/s = 11.737 m/s

viy = vi sin theta = 25sin62 m/s = 22.074 m/s

b)

we know that in projectile motion the horizonal component of the velocity never change which is constant and the vertical component of velocity only changes so

the acceleration components are a(x) = 0 m/s2 and a(y) = -g = -9.8 m/s2

c) we know that when the rock reaches the highest point the vertical component of velocity become zero but horizontal componnet of velocity remains same

so the speed at highest point is V = sqrt(vix^2+viy^2) = sqrt(11.737^2+0^2)= 11.737 m/s

speed is v = 11.737 m/s

d)

from equations of motion  

y -y0 = viy*t -0.5*gt^2

y = y0 +viy*t -0.5*gt^2

y = 0 + 22.074*3.6 - 0.5*9.8*3.6^2 m

y =15.9624 m

the height of the tank is h = 15.9624 m

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