Question #135060

A stone is thrown horizontally with an initial speed of 30 m/s from a bridge. Find the stone’s total speed when it enters the water 4 seconds later. (Ignore air resistance.)

(A) 30 m/s (B) 40 m/s (C) 50 m/s (D) 60 m/s (E) 70 m/s 


1
Expert's answer
2020-09-28T08:00:07-0400

Let's choose the upwards as the positive direction and find vertical (yy-component) of the stone's speed from the kinematic equation:


vy=v0y+gt,v_y = v_{0y} +gt,

here, v0y=0 ms1v_{0y} = 0 \ ms^{-1} is the initial vertical speed of the stone, g=10ms2g = -10 ms^{-2} is the acceleration due to gravity, t=4 st = 4 \ s is the time during which the stone flies from the bridge to the water.

Then, we get:


vy=10 ms24 s=40 ms1.v_y = -10 \ ms^{-2} \cdot 4 \ s = -40 \ ms^{-1}.

The sign minus indicates that the vertical component of the stone's speed directed downward.

Finally, we can find the stone’s total speed from the Pythagorean theorem:


v=vx2+vy2,v = \sqrt{v_x^2+v_y^2},

here, vx=30 ms1v_x = 30 \ ms^{-1} is the horizontal component of the stone's speed.

Then, we get:


v=(30 ms1)2+(40 ms1)2=50 ms1.v = \sqrt{(30 \ ms^{-1})^2+(-40 \ ms^{-1})^2} = 50 \ ms^{-1}.

Answer:

(C) v=50 ms1.v = 50 \ ms^{-1}.


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