Using Again The Above Mentioned Position Function For Free-Falling Bodies, , , Find The Instantaneous Velocity Of An Object Thown Upwards From A Build

Using Again The Above Mentioned Position Function For Free-Falling Bodies, , , Find The Instantaneous Velocity Of An Object Thown Upwards From A Build

Using again the above mentioned position function for free-falling bodies

$(y = \frac{1}{2} gt {}^{2} + v0t + y0)$
, find the instantaneous velocity of an object thown upwards from a building 95 ft high with an initial velocity of 75 ft/s.

Answer:

The instantaneous velocity is approximately at 33.03 m/s and 2.33 seconds the time it took.

Step-by-step explanation:

Given:

$y = 95 ft = 28.956 m$

$V_{y0} = 75 ft = 22.86 m$

$a_{y} = -9.81 \frac{m}{s^{2}}$

Formula:

Using this constant-acceleration equation $y = V_{y0} t + \frac{1}{2} a_{y} t^{2}$ and the quadratic formula $\frac{-b+\sqrt{b^{2}-4ac } }{2a} ; \frac{-b-\sqrt{b^{2}-4ac } }{2a}$ , by getting the positive answer only, we can find the time it took.

Using this constant-acceleration equation $(V_{yf}) ^{2} = (V_{y0}) ^{2} - 2 a_{y} y$ , we can find its final velocity or the instantaneous velocity.

Equation:

$y = V_{y0} t + \frac{1}{2} a_{y} t^{2}$

$\frac{-b+\sqrt{b^{2}-4ac } }{2a} ; \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

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$(V_{yf}) ^{2} = (V_{y0}) ^{2} - 2 a_{y} y$

$V_{yf} = \sqrt{(V_{y0}) ^{2} - 2 a_{y} y}$

Solution:

$y = V_{y0} t + \frac{1}{2} a_{y} t^{2}$

$28.956 m = 22.86 m t + \frac{1}{2} -9.81 \frac{m}{s^{2}} t^{2}$

$28.956 m = 22.86 m t - 4.905 \frac{m}{s^{2}} t^{2}$

$0 = -28.956 m + 22.86 m t - 4.905 \frac{m}{s^{2}} t^{2}$

$(0 = -28.956 m + 22.86 m t - 4.905 \frac{m}{s^{2}} t^{2}) (-1)$

$0 = 28.956 m - 22.86 m t + 4.905 \frac{m}{s^{2}} t^{2}$

where:

a = $4.905 \frac{m}{s^{2}} t^{2}$ = 4.905

b = $-22.86 m t$ = - 22.86

c = $28.956 m$ = 28.956

$t = 2.330275229 s$

$(V_{yf}) ^{2} = (V_{y0}) ^{2} - 2 a_{y} y$

$V_{yf} = \sqrt{(V_{y0}) ^{2} - 2 a_{y} y}$

$V_{yf} = \sqrt{(22.86 m) ^{2} - (2) (-9.81 \frac{m}{s^{2}}) (28.956 m)}$

$V_{yf} = 33.02569182 \frac{m}{s^{2}}$

Answer:

$t = 2.330275229 s$

$V_{yf} = 33.02569182 \frac{m}{s^{2}}$

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