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3 years ago

What height (displacement) will a ball reach if thrown upward with an initial velocity of 15 m/s

Physics

1 answer:

olchik [2.2K]3 years ago

8 0

Answer:

11.48 m

Explanation:

initial velocity (u) = 15 m/s

acceleration due to gravity (g) = 9.8 m/s

final velocity (v) = 0 ( the ball's speed reduces gradually when thrown until

becomes 0 at maximum height)

find the height reached.

height (s) = ut + 0.5 at^{2}

we cannot apply this formula directly because we do no know the time

(t), so we first need to find the time

v = u + at

0 = 15 + (-9.8 x t) (the negative sign is because the ball is decelerating

upwards)

15 = 9.8t

t = 1.53 s

now that we have the time we can put it into the initial equation

height (s) = ut + 0.5 at^{2}

height (s) = (15 x 1.53) + (0.5 x (-9.8) x 1.53^{2})

height = 22.95 - 11.47 = 11.48 m

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The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.

Explanation:

Given;

vertical height of the cliff, h = 22 m

velocity of the ball, u = 18 m/s at an angle 39⁰

vertical component of the velocity, $u_y = u \ sin \theta$

The time for the ball to get to the bottom of the cliff is calculated as;

h = ut + ¹/₂gt²

h = (u sinθ)t + ¹/₂ x 9.8 x t²

22 = (18 sin 39)t + 4.9t²

22 = 11.328t + 4.9t²

4.9t² + 11.328t - 22 = 0

Solve the above equation with formula method;

a = 4.9, b = 11.328, c = -22

$t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-11.328\ \ +/- \ \ \sqrt{(-11.328)^2 - 4(4.9\times -22)} }{2(4.9)}\\\\t = 1.26 \ s$

The time for the player to get to the bottom of the cliff;

Given maximum speed, Vx = 6.0 m/s and horizontal distance, X = 12 m;

$t = \frac{X}{V_x} \\\\t = \frac{12}{6} \\\\t = 2 \ s$

The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.

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