An Ahtlete Throws A Shot At An Angle Of 45 Degree

Anonymous

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An ahtlete throws a shot at an angle of 45 degree to the horizontal at an initial speed of 43ft/sec. it leaves his hand 7ft above the ground. where does the shot land?

 

Answer No.1

AlphaAsian3

From: US
Posts: 18
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QuestionDetails: an ahtlete throws a shot at an angle of 45 degree to the horizontalat an initial speed of 43ft/sec. it leaves his hand 7ft above theground. where does the shot land? FORMULAS v=u+gt s=ut+gt2/2 v2-u2=2gs u=initial velocity.......fps v=final velocity....fps g=32 fps2 t= time ...s VELOCITY=43 FPS VERTICAL COMPONENT=VSIN(45)=43/√2=30.4 HORIZONTAL COMPONENT=VCOS(45)=43/√2=30.4 UP WARD VERTICAL MOTION INITIAL VELOCITY=43/√2=[(43√2)/2]=21.5√2=30.4 FINAL VELOCITY=0 0=21.5√2-32T T=21.5√2/32 = 0.95 SEC..................................1 VERTICAL DISTANCE TRAVELED = 30.4*0.95-16*0.95*0.95=14.45 FT. INITIAL HEIGHT ABOVE GROUND = 7 FT. DOWN WARD VERTICAL MOTION DISTANCE OF FALL = 14.45+7=21.45 FT 21.45=0T+16T2...... T= 1.16SEC.........................2 TOTAL TIME OF TRAVEL = 0.95+1.16=2.11 SEC. HORIZONTAL MOTION VELOCITY=30.4 TIME=2.11 SEC. DISTANCE TRAVELED = 30.4*2.11 = 64.1 FT THE SHOT LANDS AT A POINT 64.1 FT AWAY HORIZONTALLY FROM WHERE HEHAS THROWN PLEASE CHECK THE ARITHMETIC ALWAYS FOLLOWING THE METHOD.

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