Question

answer plz.................☺☺☺
if the horizontal range of projectile is 4 times its maximum height the angle of projection is:

Answer 1

Step 1: write out the expression for Height (max) in terms of V (initial speed of projectile) and theta (direction of the initial speed) and “g”. [V (sin k)]^2 / 2 g

Step 2: repeat Step (1) for RANGE. [(V)^2 (sin 2 k)] / g

Step 3: multiply expression in Step (1) by a factor of 4.

Step 4: equate expressions in (2) & (3).

Step 5; use a trig identity to eliminate the term (sin 2 k). I suggest you use:

sin (2 k) = 2 (sin k) (cos k).

Step 6: after substitution, simplify the equality in Step (4).

Step 7: the final result would be: Tan k = 1/2 → k = arctan 1/2 = 26.6 degrees.

Answer 2

H =(u*sin theta) ^2/2g
R = u^2*sin 2theta /g
R =4H
u^2*2*sin theta *cos theta /g =4*u^2*sin^2/2g
Sin theta =cos theta
Theta =45