Question #201476

2. AN OBJECT IS THROWN WITH A SPEED OF 60M/SEC IN A DIRECTION 60 O ABOVE THE HORIZONTAL. FIND;

A. THE HEIGHT IT RISES

B. THE TOTAL TIME OF FLIGHT

C. THE HORIZONTAL RANGE.


1
Expert's answer
2021-06-01T10:50:09-0400

a)


h=602sin2602(9.8)=138 mh=\frac{60^2\sin^2{60}}{2(9.8)}=138\ m

b)


t=260sin60(9.8)=10.6 st=2\frac{60\sin{60}}{(9.8)}=10.6\ s

c)


R=602sin120(9.8)=318 mR=\frac{60^2\sin{120}}{(9.8)}=318\ m


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