If an object is projected horizontally from a height of 5 m with an initial velocity of 7 m/s, what is the value of x0?

Physics

1 answer:

posledela3 years ago

6 0

The horizontal displacement of the object is 3.64 m.

The given parameter:

the height of the projection, h = 5 m

initial velocity of the projection, v₀ = 7 m/s

To find:

the horizontal displacement ( $X_0$ ) of the object

The horizontal displacement of the object is calculated as;

$X_0 = v_0t$

where;

t is the time of the motion

The time of the motion is calculated as;

$h = v_0t + \frac{1}{2} gt^2\\\\5 = 7t + (0.5\times 9.8 t^2)\\\\5 = 7t + 4.9t^2\\\\4.9t^2 +7t - 5 = 0\\\\solve \ quadratic \ equation \ using \ formula \ method\\\\a = 4.9, \ b = 7 \ c = -5\\\\t = \frac{-b \ \\\ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-7 \ \ +/- \ \ \sqrt{(7^2) - (4 \times 4.9 \times -5)} }{2(4.9)} \\\\t = \frac{-7 \ + /- \ 12.12 }{9.8} \\\\t = \frac{-7 + 12.12}{9.8} \\\\t = 0.52 \ s$

The horizontal displacement of the object is calculated as;

$X_0 = v_0 t\\\\X_0 = 7 \times 0.52\\\\X_0 = 3.64 \ m$

Thus, the horizontal displacement of the object is 3.64 m.

Learn more here: brainly.com/question/4582453

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