"AST1CAL3 EQUATION VARIABLE","11-07-1993","12:08:45" "VELOCITY=(G*T)*(1-SIGN(ABS(V)))+SIGN(ABS(V))*V HEIGHT=(0.5*VELOCITY*T)*(1-SIGN(ABS(H)))+SIGN(ABS(H))*H TIME=(SQR(ABS(2*HEIGHT/G)))*(1-SIGN(ABS(T)))+SIGN(ABS(T))*T" "UNIFORM ACCELERATED MOTION in a VERTICAL PLANE. Falling bodies, near the earth, all have a uniform downward acceleration from the earth's gravity, G=32 ft/s^2 or 9.8 m/s^2. A uniform gravitation field G will result in VELOCITY at TIME and a distance, or HEIGHT, for the free fall. start at rest, initial velocity=0 | ----- G ³ V ³ gravity³ | HEIGHT V V ³ <- VELOCITY at TIME ----------- Use these equations when velocity and time are important factors. G=gravity, V=final velocity at time T, H=distance traveled at time T. Enter value of gravity G and at least two other variables with nonzero values. (c) Copyright PCSCC, Inc., 1993 *** Answer to problem *** Type ? then (esc) 0 (enter) to set all variables to zero. Set the gravity to 32ft/s. Move the cursor to T and set the time to 3.95 s. The velocity at impact with the water is 126.4 ft/s and the bridge is 249.6 ft high. Type any key to exit. ||A superball dropped from the rivergate bridge hits the water in 3.95 s. (a) What is its final velocity in ft/s and (b) how high is the rivergate bridge above the river? Type comma key to see answer. Type (F2) to return to helpfile." 7 126.4,0,"" 249.64,0,"" 3.95,0,"" 32,0,"" 3.95,0,"" 0,0,"" 0,0,"" 1 0 0