A Ball Was Thrown From The Ground At A Speed Of 20 M/S At 30 Deg. Determine The Horizontal And Vertical Displacements Traveled By The Ball At 1.25 S.

A ball was thrown from the ground at a speed of 20 m/s at 30 deg. Determine the horizontal and vertical displacements traveled by the ball at 1.25 s.

Answer:

The horizontal and vertical displacements travelled are approximately at x = 12 m and y = -32.

Explanation:

Given:

$V_{0}$ = 20 m/s

a = 9.81 $\frac{m}{s^{2}}$

θ = 30°

t = 1.25 s

Formula:

For horizontal x-component ( $V_{0x}$ ): $V_{0}$ cosθ

For vertical y-component ( $V_{0y}$ ): $V_{0}$ sinθ

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For horizontal displacement (x): $x = V_{0x}t + \frac{1}{2}a_{x} t^{2}$

For vertical displacement (y): $y = V_{0y}t - \frac{1}{2}a_{y} t^{2}$

Solution:

Horizontal x-component: (20 m/s)(cos 30°)

$V_{0x}$ = 3.085028998 m/s

Vertical y-component: (20 m/s)(sin 30°)

$V_{0y}$ = -19.76063248 m/s

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Horizontal displacement:

$x = V_{0x}t + \frac{1}{2}a_{x} t^{2}$

$x = (3.085028998 m/s)(1.25 s) + \frac{1}{2}(9.81 \frac{m}{s^{2}})(1.25 s)^{2}$

x = 11.52034875 m

x ≈ 12 m (**NOTE: Positive means to the right.)

Vertical displacement:

$y = V_{0y}t - \frac{1}{2}a_{y} t^{2}$

$y = (-19.76063248 m/s)(1.25 s) - \frac{1}{2}(9.81 \frac{m}{s^{2}})(1.25 s)^{2}$

y = -32.3648531 m

y ≈ -32 m (**NOTE: Negative means going down.)

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