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Gnesinka [82]
3 years ago
13

3. How long will it take a ball to reach the ground if it falls from a cliff 25 m tall (starting from rest)? What if the ball we

re given an initial horizontal velocity of 5 m/s? please help,thanks​
Physics
1 answer:
Nady [450]3 years ago
4 0

Answer:

2.26 s

Explanation:

Let's take down to be positive.

Given (in the y direction):

Δy = 25 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

25 m = (0 m/s) t + ½ (9.8 m/s²) t²

25 = 4.9t²

t = 2.26 s

If the ball instead had an initial horizontal velocity of 5 m/s, its initial vertical velocity is still 0 m/s.  So the time to fall is still 2.26 s.

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0.0000000010 joules

Explanation:

Amount : 1540.3 nanojoules (nJ)

Equals : 0.0 joules (J)

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Which organism would most likely have new gene combinations?
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Who is the greek philisopher who came up with the idea that atoms exist
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4 0
3 years ago
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
algol13

Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
  • Vertical: constant downward acceleration at g = 10\; \rm m \cdot s^{-2}, starting at 0\; \rm m \cdot s^{-1}.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

5 0
3 years ago
HI PLEASE HELP, I ONLY HAVE 10 MINUTES
Whitepunk [10]

Answer:

um ok

?????

i dont like you get rejected

jk u gud

Explanation:

5 0
3 years ago
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