pA 2.69 kg ball is dropped from the roof of a building 161.1 m high. While the ball is falling to Earth, a horizontal wind exerts a constant force of 12.7 N on the ball. How long does it take to hit the ground? The acceleration of gravity is 9.81 m/s2 .
+6251 Assuming the ground is flat everywhere the horizontal wind won't affect the time it takes for the ball to fall to the ground. The mass of the ball has no effect on the that time either.
\(h(t)=-\dfrac 1 2 g t^2 + v_0 t + h_0\\ h_0=161.1m \\ v_0 = 0 m/s \\ g=9.81 m/s^2 \\ \mbox{plug all that in and solve }h(t)=0 \mbox{ for }t \\ \mbox{You'll get two solutions. One will make sense.}\)
In this situation, the horizontal wind is just a distraction.....it does nothing to affect the time it takes for the ball to hit the ground once it is dropped; The same goes for the mass of the ball: no effect
p = po +vt + 1/2 at^2 the position when it hits the ground will be 0 po = original position =161.1 m v=velocity= 0 a = -9.81
0=161.1 + 1/2 (-9.81)t^2
161.1/(1/2) = 9.81 t^2
322.2/9.81 = t ^2
5.73 = t
+6251 Assuming the ground is flat everywhere the horizontal wind won't affect the time it takes for the ball to fall to the ground. The mass of the ball has no effect on the that time either.
\(h(t)=-\dfrac 1 2 g t^2 + v_0 t + h_0\\ h_0=161.1m \\ v_0 = 0 m/s \\ g=9.81 m/s^2 \\ \mbox{plug all that in and solve }h(t)=0 \mbox{ for }t \\ \mbox{You'll get two solutions. One will make sense.}\)
