Question

physics

Suppose the stone is thrown at an angle f 40.0° below the o orizontal from the same b uilding (h 42.0 m) as in the example above. If it strikes the ground 38.9 m away, find the following. (Hint: For part (a), use the equation for the x displacement to eliminate vot from the equation for the y displacement.) (a) the time of flight (b) the initial speed m/s (c) the speed and angle of the velocity vector with respect to the horizontal at impact speed m/s angle o below the horizontal

Answer 1

Answer 2

We will start with finding the x component equation

In the x, d = vt

38.9 = (v)(cos 40)(t)

t = 50.78/v

In the y direction

d = vot + .5at²                

42 = vsin(40)(50.78)/v + .5(9.8)(50.78²/v²)

42 = 32.6 + 12635/v²

v = 36.7 m/s (Part B Answer)

Thus t = 50.78/36.7

t = 1.38 sec (Part A Answer)

Next we need the final velocity in the y direction

vf = vo + at

vf = (30.67)(sin 40) + (9.8)(1.38)

v = 33.2 m/s

The net speed is from the pythagorean theorem

v2 = (33.2)² + [(36.7)(cos 40)]²

v = 43.5 m/s

Angle from the tangent function

tan(angle) = 33.2/(36.7)(cos 40)

angle = 49.7^o below the horizontal