Question #265402

A ball is thrown horizontally from a building 63.0 m high with a speed of 25.0 m/s. Find the (a)vertical and horizontal components of the ball's initial velocity, (b) time flight, (c) distance from the foot of the building where the ball will strike the ground, and (d) velocity when the ball will strike the ground.


Expert's answer

(a) vx0=25 (m/s)v_{x0}=25\ (m/s) and vy0=0v_{y0}=0


(b) h=gt2/2t=2h/g=263/9.8=3.6 (s)h=gt^2/2\to t=\sqrt{2h/g}=\sqrt{2\cdot63/9.8}=3.6\ (s)


(c) x=vxt=253.6=90 (m)x=v_x\cdot t=25\cdot3.6=90\ (m)


(d) v=vx2+vy2=vx2+(gt)2=v=\sqrt{v_x^2+v_y^2}=\sqrt{v_x^2+(gt)^2}=


=252+(9.83.6)2=43.3 (m/s)=\sqrt{25^2+(9.8\cdot 3.6)^2}=43.3\ (m/s)






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