Question

Ahiker throws a ball at an angle of 21.0" above the horizontal frornョcliff 21 magnitude of the of the velocity is 15.000 m/s. Find a. The magnitudes of the horizontal and vertical componenis 3. horizontal from a citt 21. m h T m hith The of the velocity, respecty (PSYw) (4) The horizontal distance between the base of the hill and the point where the ball hits the grou (PSYW) (4) b. c. The time it took for the ball to hit the ground below. (PSYw) (4) d. The maximum height attained by the ball above the cliff (PSYW) (4)

Answer 1

(a) The magnitude of the horizontal and vertical component of velocity will be

$u_x=u_0\cos\theta=15\cos21=14.0037\\ u_y=u_0\sin\theta=15\sin21=5.3755\\$

(b) When the ball hit the ground the vertical displacement is -21m( considering the point of projection as origin)

$h=u_yt-\frac{1}{2}gt^2\Rightarrow -21=5.3755t-\frac{1}{2}\times9.8\times t^2\\ 4.9t^2-5.3755t-21=0\Rightarrow t=2.690$ and $ t=-1.593$

Ignoring negative time, hence t=2.690

Horizontal distance traveled $x=u_xt= 14.0037\times2.690=37.67$ m$$ will be

(c) Time is t=2.690 sec

(d) At maximum height, vertical velocity will be zero hence,

$0=u_y^2-2gh\Rightarrow h=\frac{u_y^2}{2g}=\frac{5.3755^2}{2\times9.8}=1.4743$ m$$

Total height will be