Question 64597: I really, really need help on this one..... The (-32x^2 over 310^2), in the equation below, is a fraction......
A projectile is fired from a cliff 200 feet above the water at an incline of 45 degrees to the horizontal, with a muzzle velocity of 310 feet per second. The height h of the projectile above the water is given by:
h(x)= (-32x^2 over 310^2)+x+200 , where x is the horizontal distance of the projectile from the base of the cliff. Find the maximun height of the projectile.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I really, really need help on this one..... The (-32x^2 over 310^2), in the equation below, is a fraction......
A projectile is fired from a cliff 200 feet above the water at an incline of 45 degrees to the horizontal, with a muzzle velocity of 310 feet per second. The height h of the projectile above the water is given by:
h(x)= (-32x^2 over 310^2)+x+200 , where x is the horizontal distance of the projectile from the base of the cliff. Find the maximun height of the projectile.
H=(-32X^2/310^2)+X+200
HORIZONTAL COMPONENT OF VELOCITY=310COS(45)=310/SQRT(2)=310SQRT(2)/2=155SQRT(2)=219.2 FT/SEC
VERTICAL COMPONENT OF VELOCITY =310SIN(45)=155SQRT(2)=219.2 FT/SEC
THE PROJECTILE WILL GO UP TILL ITS VELOCITY=0 AND THEN FALLS DOWN.
FORMULA IS V=U+AT...
WHERE U = INITIAL VELOCITY IN THE VERTICAL DIRECTION=155SQRT(2) FT/SEC
V=FINAL VELOCITY =0 FT/SEC
A=ACCELERATION DUE TO GRAVITY = -32 FT/SEC^2
T= TIME OF TRAVEL IN SEC.
0=155SQRT(2)-32T
T=155SQRT(2)/32=6.85 SEC
HORIZINTAL TRAVEL IN THIS TIME OF 6.85 SEC = 155SQRT(2)*6.85=1501.5
H=-(32*1501.5^2/310^2)+1501.5+200=950.5 FEET.ABOVE THE WATER LEVEL OR 750.5 FEET ABOVE THE CLIFF LEVEL.
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PLEASE NOTE THAT THERE ARE SEVERAL WAYS TO DO THIS PROBLEM,USING FORMULAE,CALCULUS ETC.IF I KNOW YOUR COURSE OF STUDY AND BACK GROUND ,I CAN GIVE THE METHOD BEST SUITED TO YOU.
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