Answer to Question #126095 in Classical Mechanics for Quanda

Question #126095
A vertical wall of height 3.75 m is at a horizontal distance of 5 m from a point O on the ground. A particle is projected from O with initial speed 10 m/s at an angle of elevation x. Given that the particle just clears the wall, find
A) the value of tanx
B) the speed of the particle as it passes over the wall
C) the horizontal distance of the particle from the wall when it again reaches the level of O. Take g as 10 m s^-2
1
Expert's answer
2020-07-13T11:42:29-0400

A). since,

"Y=\\tan(x)X+\\frac{gX^2}{2u^2}(1+\\tan^2(x))"

Now,

"3.75=5\\tan(x)+\\frac{5}{4}(1+\\tan^2(x))"

Let,"p=\\tan(x)" ,thus

"p^2+p-2=0\\implies (p+2)(p-1)=0\\\\\\implies p=1,-2"

By convention, since "x" is positive and "0^\\circ<x\\leq 90^\\circ" ,hence "p=1\\iff \\tan(x)=1"


B).As "5=u\\cos(x)t\\implies t=\\frac{1}{\\sqrt{2}}" and "v_y=u\\sin(x)-gt\\implies v_y=0" ,thus

"v=\\sqrt{v_x^2+v_y^2}=5\\sqrt{2}m\/s"



C).When the particle reaches the ground

"3.75=\\frac{1}{2}gt^2\\implies t=\\frac{\\sqrt{3}}{2}"

And,

"x_1=u\\cos(x)t=5\\sqrt{\\frac{3}{2}}\\approx6.12m"


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 the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS