Question

A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starting point of the ball, and whose y axis points vertically upward. Neglecting air resistance, calculate the ve- locity of the ball after 3.04 s. The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s.

Answer 1

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: $Vf^{2} = Vo^{2} - 2gh$ and $h=Vo*t - \frac{g*t^{2} }{2}$, First we need to find the height to time iqual to 3.04 s using the formula: $h=Vo*t - \frac{g*t^{2} }{2}$ and remember that golf ball was released from the rest (Vo= 0 (m/s)) so $h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}$, we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using $Vf^{2} = Vo^{2} - 2gh$ solving for Vf, we get: $Vf = \sqrt{Vo^{2}-2*g*h }$ replacing the values given $Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }$, so we get: Vf = 29.79 (m/s).

Answer 2

Answer: $v=29.792m.s^{-1}$

Explanation:

Here given that:

• initial velocity of the golf ball, $u =0m.s^{-1}$
• a coordinate system pointing upwards with origin at the point of release.
• time, t = 3.04 seconds
• acceleration due to gravity, $g = 9.8 m.s^{-2}$

Now, we need to find the velocity (v) after the time 3.04 s.

From the given and required data we choose the suitable equation of motion: $v = u+at$

Putting the given values in the above equation:

$v= 0 - gt$ ∵the given frame of reference points in the upward directoin so we have negative on the downward side.

$v=0-9.8\times 3.04$

$v=29.792m.s^{-1}$