A ball dropped from a bridge takes three seconds to reach the water below how far is the bridge above the water?

Physics

1 answer:

tatuchka [14]3 years ago

6 0

Given that:

Ball dropped from a bridge at the rate of 3 seconds

Determine the height of fall (S) = ?

As we know that, S = ut + 1/2 ×a.t²

u =initial velocity = 0

a= g =9.81 m/s (since free fall)

S = 0+ 1/2 × 9.81 × 3²

S = 44.145 m

44.145 m far is the bridge from water

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valentinak56 [21]

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4 0

4 years ago

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

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4 0

2 years ago

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where U is potential energy and K is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$