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Anna007 [38]
3 years ago
7

A ball is thrown straight upward and returns to the thrower’s hand after 1.8 s in the air. A second ball is thrown at an angle o

f 23◦ with the horizontal. At what speed must the second ball be thrown so that it reaches the same maximum height as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Physics
1 answer:
zysi [14]3 years ago
5 0

Answer:

U = 9.1 m/s

Explanation:

from the question we are given the following

time (t) = 1.8 s

angle = 23 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below

v = u + 0.5at

  • The final velocity (v) is zero since the ball comes to rest
  • The time (t) it takes to get to the maximum height would  be half the time it is in the air, t = 0.5 x 1.8 = 0.9

therefore

0 = u - (0.5 x 9.8 x 0.9)

u = 7.9 m/s

for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore

U sin 60 = 7.9

U = 7.9 ÷ sin 60

U = 9.1 m/s

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The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the ele
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From the question we are told that

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    The  thickness of the membrane is  t =  9.22 \ nm  =  9.22 *10^{-9} \ m

     

Generally the electric field intensity is mathematically represented as

                E =  \frac{\epsilon }{t}

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8 0
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1. A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ba
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Momentum = (mass) x (velocity)

Original momentum before the hit =

                   (0.16 kg) x (38 m/s) this way <==

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Change in momentum = impulse.

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                          13.12 kg-m/s  = (force) x (0.002 sec)

  (13.12 kg-m/s) / (0.002 sec)  =  Force

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Answer:

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