Question

A ball is thrown at an angle x and another ball is thrown at angle (90°-x) with horizontal direction from the same point with velocity 40 m/s. The second ball reaches 50 m higher than the first ball. Find their individual heights. g=10m/s2

Answer 1

By newton equation of motion, v²=u²-2as ------------(1)

For the first ball, final velocity(v) is zero and initial velocity(u) is 40 sinx m/s.

In (1) , 0=(40sinx)²-2*10*s

1600sin²x=20s ------------------------(2)

For the seond ball, v=0 and u=40sin(90-x) m/s

Putting in (1), 0=(40sin(90-x))²-2*10*(s+50)

1600sin²(90-x)=20s+1000 -----------------------(3)

Adding (2) and (3)

1600sin²(x+90-x)=20s+20s+1000

1600sin²90=40s+1000

1600-1000=40s (sin 90=1)

s=600/40

s=15m

s+50=65m