A. plasmodium
D. plague
E.yellow ever
F. Typhus
(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
Answer:
The time it takes the stone to reach the bottom of the cliff is approximately 4.293 s
Explanation:
The given parameters are;
The height of the cliff, h = 90.4 m
The direction in which the stone is thrown = Horizontally
The speed of the stone in the horizontal direction = 10 m/s
The time, t, it takes the stone to reach the bottom of the cliff is given by the equation for free fall as follows;
h = 1/2 × g × t²
Where;
g = The acceleration due to gravity = 9.81 m/s²
Substituting the values gives;
90.4 = 1/2 × 9.81 × t²
t² = 90.4/(1/2 × 9.81) ≈ 18.43 s²
t = √18.43 ≈ 4.293 s
The time it takes the stone to reach the bottom of the cliff is t ≈ 4.293 s.
The answer you're looking for is True