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3 years ago

A kid throws a ball from a tower 45 meters tall how fast is the ball going when it hits the ground?

Physics

1 answer:

IRISSAK [1]3 years ago

8 0

Answer:

29.69 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) = 45 m

Velocity (v) =...?

Next, we shall determine the time taken for the ball to get to the ground. This can be obtained as follow:

Height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.9 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the velocity with which the ball hits the ground. This is illustrated below:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 3.03 s

Final velocity (v) =.?

v = u + gt

v = 0 + (9.8 × 3.03)

v = 0 + 29.694

v = 29.694 ≈ 29.69 m/s

Therefore, the ball hits the ground with a velocity of 29.69 m/s.

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2 years ago

A plane drops a rubber raft to the survivors of a shipwreck. The plane is flying at a height of 1960 m and with a speed of 109 m

Korolek [52]

In this case the rubber raft has horizontal and vertical motion.

Considering vertical motion first.

We have displacement $s = ut +\frac{1}{2}at^2$ , u = Initial velocity, t = time taken, a = acceleration.

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s = 1960 m, u = 0 m/s, a = 9.81 $m/s^2$

$1960 = 0*t+\frac{1}{2} *9.81*t^2\\ \\ t = 20 seconds$

So raft will take 20 seconds to reach ground.

Now considering horizontal motion of raft

u = 109 m/s, t = 20 s, a = 0 $m/s^2$

So $s = 109*20+\frac{1}{2} *0*20^2 = 2180 m$

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2 years ago

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$