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A particle is launched from the top of a cliff of height 87.5m at time t=0 with initial velocity 14m/s at 30 deg above the horizontal, Calculate: a) maximum height reached above bottom of cliff; b)horizontal distance travelled before hitting the ground.

For parabolic motion: vertical and horizontal components can be treated independently. x represents horizontal motion, which obeys constant velocity equations, y denotes vertical motion, which obeys the cons...
JK
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A=[5k,3k-1;-3,k+1] where k is a real constant. Given that A is singular, find all the possible values of k.

0.2, -3
DB
4108 Views

What is the value of x from (x+2)^2=4

(x+2)^2=4=>x+2=2+>x=0 or x+2=-2=>x=-4
AB
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By forming and solving a suitable quadratic equation, find the solutions of the equation: 3cos(2A)-5cos(A)+2=0

3cos(2A)-5cos(A)+2=0. The first thing we do is use a trignometric identity, namely cos(2A) = 2(cos^2(A))-1. This gives us a new form of the original equation. 3(2(cos^2(A))-1)-5cos(A)+2=0: we expand out the ...
JB
3711 Views

The plane Π contains the points (1, 2, 3), (0, 1, 2) and (2, 3, 0). What is the vector equation of the plane? and what is the cartesian equation of the plane?

Vector Equation So we know it contains three points so we can find two lines in the plane. 1) (1,2,3) + A((0,1,2) - (1,2,3)) = (1,2,3) + A(-1,-1,-1) 2) (1,2,3) + B((2,3,0) - (1,2,3)) = (1,2,3) + B(1,1,-3) Ge...
OO
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