A Baseball Was Hit And Projected Horizontally At Angle Of 45\Xb0. If The Initial Velocity Of The Ball Was 75 M/S (Neglect Air Friction), Calculate The

A Baseball Was Hit And Projected Horizontally At Angle Of 45\Xb0. If The Initial Velocity Of The Ball Was 75 M/S (Neglect Air Friction), Calculate The

A baseball was hit and projected horizontally at angle of 45°. if the initial velocity of the ball was 75 m/s (neglect air friction), calculate the following:

a. time the baseball was in the air

b. horizontal distance the ball traveled

c. maximum height the ball reached

$V_{0} =$ 75 m/s

∅ = 45°

a = -9.81 $m/s^{2}$

$V_{x} = V_{0}$ cos∅

$V_{y} = V_{0}$ sin∅

$R_{V} = \sqrt{(V_{y}) ^{2} + (V_{x}) ^{2}}$

∅ = tan $^{-1}$ $(\frac{V_{y} }{V_{x} } )$

$V_{0x} =$ (75 m/s)(cos45)

$V_{0x} =$ 53.03300859 m/s

$V_{0y} =$ (75 m/s)(sin45)

$V_{0y} =$ 53.03300859 m/s

$R_{V} = \sqrt{(53.03300859 m/s) ^{2} + (53.03300859 m/s) ^{2}}$

$R_{V} =$ 75 m/s

∅ = 45°

(a) For the time the baseball was in the air:

$V_{f} = V_{0} + at$

$\frac{V_{f} }{t} = V_{0} + a$

$t = \frac{V_{f} - V_{0} }{a}$

$t = \frac{0 m/s-75 m/s}{-9.81 m/s^{2} }$

t = 7.645259939 s.

(b) For the horizontal distance:

$(V_{f})^{2} = (V_{0x})^{2} + 2ax$

$\frac{1}{x} = (V_{0x})^{2} + 2a$

$x = \frac{(V_{f})^{2}-(V_{0x})^{2}}{2a}$

$x = \frac{0m/s-(53.03300859 m/s)^{2}}{2(-9.81 m/s^{2})}$

x = 143.1880684 m

(c) For the maximum height:

From $(V_{f})^{2} = (V_{0})^{2} + 2ay$ we have, since $g = 9.81$ $m/s^{2}$ in the Earth,

$0 = (V_{0})^{2} + 2ay$

$\frac{1}{y} = (V_{0})^{2} + 2a$

$y = \frac{(V_{0})^{2}}{2a}$

0 = (75 m/s) $^{2}$ + 2(-9.81 $m/s^{2}$ )y

$y = \frac{-(75 m/s)^{2}}{2(-9.81 m/s^{2})}$

y = 286.6972477 m.

Note: I am not sure with my solution. Sorry.

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