## Related questions with answers

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Solution

Verifieda) To find the air time it's enought to consider only the vertical componenet of the motion, this is just a free fall from that perspective, so we have:

$h = \frac{1}{2}gt^2$

And if we solve for $t$:

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 60 \text{ m}}{ 9.81 \text{ m/s}^2 }} = \boxed{ 3.5 \text{ s}}$

b) In the horizontal direction, the velocity is constant, so the inital speed to traverse 100 meters must've been:

$v_0 = \frac{D}{t} = \frac{100 \text{ m}}{3.5 \text{ s}} = \boxed{ 28.57 \ \dfrac{m}{s} }$

c) From the arguments from a) we have that impact velocity is:

$v_\bot = gt = 9.81 \text{ m/s}^2 \cdot 3.5 \text{ s} = \boxed{ 34.33 \ \dfrac{m}{s} }$

d) To find the magnitude of the total impact velocity we use the pythagorean theorem:

$v_{tot} = \sqrt{v_0^2 + v_\bot^2} = \sqrt{(28.57 \text{ m/s})^2 + (34.33 \text{ m/s})^2} = \boxed{ 44.66 \ \dfrac{m}{s} }$

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