Question

45 (2 pts) A stone thrown off a bridge 20 m above a river has an initial velocity of 12 m/s at an angle of 45 o above the horizontal, as shown in Figure. a. What is the range R of the stone? b. Find magnitude and direction of the velocity of the stone when it strikes the water. (ANS. to 4 significant digits.) Range R- Magnitude: m/s Direction: deg

Answer 1

Vertical speed implies 12 sin 45 = 8.485 m/s time to reach at ground s = ut + 1/2 at^2 -20 = 8.485 t - 4.gt^2 t = 3.064 second Range = speed times time R = 8.484 times 3.064 (a) R = 25.9g meter Final velocity implies in x rightarrow velocity would be same v_x = 8.485 m/s v^2 = u^2 + 2as v^2_y = 72 - 2 times 9.81 times (-20) v_y = -21.5499 m/s (b) v = squareroot vx^2 + uy^2 v = 23.1603 m/s theta implies tan^-1 (21.5499/8.485) = 68.51 degree below the)