Question

A particle is projected with a velocity of 60 m/s at an angle of elevation tan-1 4/3 after 3 sec it will move at an angle a alpha to the verticall such that tan alpha equals to

Answer 1

Answer:

Explanation:

write equation of trajectory

y=xtantheta -g/2 ( x^2 / u^2cos^2 theta )

dy/dx = tan theta -gx / u^2cos^2 theta

tan alpha = tan theta - gx/ u^2cos^2 theta

tan alpha = 4/3 - 10x / 3600(cos^2 theta )

theta = 53 degrees

cos 53 = 3/5

cos^2 53 = 9/25

tan alpha = 4/3 - 10x/3600(9/25 )

tan alpha = 4/3 - 250x/3600*9

Sx = uxt

Sx = (60cos53)3 = 60(3/5)3 =108 m

tan alpha = 4/3 - 250*108 / 3600*9

tan alpha = 4/3 - 0.834

tan alpha = 1.5/3

tan alpha = 1/2

alpha = tan^-1 ( 1/2)