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Gennadij [26K]
3 years ago
12

A unique type of basketball is played on the planet zarth. During the game, a player flies above the basket and drops the ball i

n from a height of 10 m. If the ball takes 5.0 s to fall, find the acceleration due to gravity on zarth.
Physics
2 answers:
pickupchik [31]3 years ago
8 0

Answer:

The acceleration due to gravity on zarth = 0.8m/s^2

Explanation:

The acceleration due to gravity on earth = 9.8m/s^2

We know that time taken(t) by an object to fall a distance (d) is given by

t = \sqrt{ (2d/g)}

We are given the height from which the ball is dropped = 10m and the time taken by the ball to fall = 5.0s

So, substituting the given values

d = height of dropping the ball = 10m, t = 5.0 s

i.e. time taken for fall ,in the equation,

t^2 = 2d/g

g= 20/5^2 = 20/25 = 0.8

Therefore (g) = 0.8m/s^2 which gives the acceleration due to gravity on planet Zarth.

Murljashka [212]3 years ago
6 0

Answer:

0.8 m/s^2

Explanation:

The motion of an object in free fall is a uniformly accelerated motion, so we can analyze it using the following suvat equation:

s=ut+\frac{1}{2}gt^2

where, taking downward as positive direction:

s is the vertical displacement

u is the initial velocity

t is the time of flight

g is the acceleration of gravity

For the ball dropped in this problem,

s = 10 m

t = 5.0 s

u = 0

Therefore, solving the equation for g, we find the acceleration due to gravity:

g=\frac{2s}{t^2}=\frac{2(10)}{5^2}=0.8 m/s^2

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6 0
3 years ago
Read 2 more answers
A stationary block resting on the ground is pulled up with a tension force of 100N, but does not leave the ground.
tekilochka [14]

Answer:

The normal force the ground exerts on the block, F = -300 N

Explanation:

Given data,

The block pulled up with a tension force, T = 100 N

The weight of the block, W = 300 N

The weight of the block is due to the force of attraction of gravitation.

The surface exerts a force that is equal and opposite to the force acting on the block due to gravitation.

The weight of the block,

                                  W = mg

                                          300 N

The normal force the ground exerts on the block,

                                  F = - mg

                                    = - 300 N

Hence, the normal force the ground exerts on the block, F = -300 N

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The boat's position x relative to its starting point x_0=0 is determined by

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x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2

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