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Natalka [10]
3 years ago
13

g John is walking along a trail when he comes to the bottom of a steep cliff. Before trying to climb up it, he wonders how high

it is. Jake happens to have a bow and arrow with him, and he knows it is able to launch an arrow with a maximum speed of 60.0 m/s. After shooting the arrow straight up with the bow fully extended, he hears it hit the ground at the top of the cliff 7.00 s after he launches it. How high is the cliff
Physics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

179.655m

Explanation:

Given

Maximum speed of the arrow v = 60m/s

Time taken to hit the top of the cliff t = 7.0s

Required

Height of the cliff H

Using the equation of motion

H = vt + 1/2gt²

Substitute into the formula:

H = 60(7) + 1/2 (-9.81)(7²) (g is negative due to upward motion of the arrow)

H = 420-4.905(49)

H = 420-240.345

H = 179.655m

Hence the cliff is 179.655m high

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Answer:

<h2>73 kg</h2>

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The mass of the object can be found by using the formula

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From the question we have

m =  \frac{219}{3}  \\

We have the final answer as

<h3>73 kg</h3>

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a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

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3 years ago
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