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aniked [119]
3 years ago
10

An object is thrown directly upwards from the ground at a velocity of 9ms. Recalling that the acceleration due to gravity is −gm

s2, find the time at which the object reaches its maximum height. What is this maximum height? Your answers should be in terms of g.
Physics
1 answer:
olga55 [171]3 years ago
6 0

Answer:

Explanation:

Given

Object is thrown with a velocity of u=9\ m/s

Acceleration due to gravity is -g (i.e. acting downward)

Vertical distance traveled by object is given by

v^2-u^2=2as  

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

at maximum height final velocity is zero

0-(9)^2=2\times (-g)\times (s)

s=\frac{81}{2g}=\frac{40.5}{g}\ m

time taken to reach maximum height

using

v=u+at

0=9-gt

t=\frac{9}{g}\ s

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aniked [119]
<span>You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
 
Need to find xf, knowing xi = 0

Thus, use the formula
 x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
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20m/s = xm/20s 20*20 = x
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thus we have 100+400 = 500m then it slows down from 500m to x_f
 
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
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</span>
<span>to calculate average velocity
550/35 = 16m/s

thus V_xavg = 16m/s</span>
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