Question

6. Someone kicks a ball off of a 15-m-high cliff. The kick is at an angle of 30 degrees above the horizontal with a speed of 20 m/s. In this problem you will be ultimately solving for the distance from the cliff that the ball lands. (a) Write down all the given information. Be sure to organize it effectively. (b) Solve for the time it will take the ball to hit the ground. (c) Solve for the horizontal distance from the cliff that the ball will land.

Answer 1

Given: Initial launch velocity,

u= 20m/s at 30° above horizontal

Initial vertical velocity,

uy = usin0 = 20m/s * sin30 = 10m/s

Initial horizontal velocity,

uz = ucosé = 20m/s * cos30 = 17.32m/s

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Consider the vertical motion of the ball

Consider the rising part

Use formula

U=u+ at

0gmax = uy-gt-

At maximum height, vertical velocity becomes zero. Put a negative sign for g because the ball is going up and g is acting downwards.

Om/s = 10m/s - 9.81m/s2 * t,

0 = 10 - 9.81 * to

10 = 9.81 *t

10 9.81 =tr

ty = 1.019s

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Use formula

2-u= 295

vymar – u = -2gh

At maximum height, vertical velocity becomes zero. Put a negative sign for g because the ball is going up and g is acting downwards.

(Om/s) – (10m/s) = -2 * 9.81m/s-*h

0% - 10% = -2 *9.81 *h

- 10% = -2*9.81 *h

10% = 2 * 9.81 * h

102 h 2* 9.81

1 = 5.lm

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Maximum Height = Cliff Height + Height reached by the ball

H = họ + h = 15m +5.lm

H = 20.lm

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Consider the falling part

Use formula

s= ut + -at-

H = Vymar * tp +=gt;

20.1m = 0m/s *t +0.5*9.81m/s*t?

20.1 = 0*tf + 0.5 * 9.81*t*

20.1 = 0.5 * 9.81 *t*

20.1 V 0.5*9.81 = ty

ty = 2.024s

=================

Time of flight,

t=t, + y = 1.019s + 2.024s = 3.043s

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Consider the horizontal motion of the ball

There is no acceleration in horizontal direction, so horizontal velocity remains constant.

Horizontal velocity = Horizontal distance/Time of flight

داد

U *t=

17.32m/s *3.043s = 1

ANSWER:
r = 52.7m