In Fig. 4-34, a stone is projected at a cliff of height h with an initial speed
ID: 1695884 • Letter: I
Question
In Fig. 4-34, a stone is projected at a cliff of height h with an initial speed of 49.0 m/s directed at an angle 0 = 63.0° above the horizontal. The stone strikes at A, 5.54 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.
Fig. 4-34
Problem 28.
Explanation / Answer
Let Initial speed, U = 49 m/ s Angle of projection, ? = 63 degree time of travel , t = 5.54 s Horizontal component of initialvelocity, Ux = U cos ? = 49* cos 63 = 22.24 m / s Vertical component of initial velocity, Uy = U sin ? = 49 * sin63 = 43.65 m / s (a) S = Ut + ( 1/2 ) a t2 h = Uy t - ( 1/2 ) g t2 = ( 43.65 * 5.5 4) - ( 4.9 * 5.54^2 ) = 241.821- 150.388 = 90.97 m (b) Speed of the stone just beforeimpact, V = U + at = Uy - g t = 43.65 - ( 9.8 * 5.54 ) = - 10.642 m/s ( downward direction ) (c) Maximum height above the ground, H =( Uy ) ^2 / 2 g = ( 43.65 )^ 2 / 2*9.8 = 97.21 m