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The more questions answered the more points that I will give. I need to have these done but im unsure of the steps that I need to take. pleas do everything step by step so I can see the question and do it correctly.

A ball is thrown straight downwards from the roof of a tall building. The building is 100.0 m high, a) when does it hit the ground? b) how fast is it moving just before it hits the ground? A toy car is moving along a straight track. It passes by a sign at 15.0 m/s and just as it passes, it slows down at a rate of 0.30 m/s2, when does it stop? b) how far did it travel from the sign when it stops? A toy plane is tracking off from a flat runway with a constant acceleration. When it passes, a physics student measures its speed as 10.0 m/s and when it just starts to take-off, another student measures its speed as 28.0 m/s. If the students are 40 m from each other, what is the planes acceleration? A particle starts from rest and accelerates in the z-axis at a constant 12.0 m/s2. At what position from the origin docs it attain a speed of 62 m/s? A stone is thrown straight upwards from the ground and reaches its maximum height in 2.80 s before falling back down, a) determine the velocity just before it hits the ground, b) what is its maximum height? A policeman is traveling down a straight section of the LIE at 25.0 m/s and spies a motorist 15.0 m ahead of him traveling in a parallel direction on the side service road. At the same time the motorist, who was initially traveling at 38.0 m/ s, sees the policeman in his side mirror, and fearing that he will get a ticket, immediately brakes at a constant rate of 1.00 m/s2, a) how long does it take for the policemen to overtake the motorist b) what is the speed of the motorist when this happens?

Explanation / Answer

1)
given data:
height h=100m
a=g(acceleration due to gravity)=9.8m/s^2
intial velocity u=0m/s

a) using equation of motion
h=u*t+0.5 *a*t^2

substututing the given values

100=0*t+0.5*9.8*t^2
=>100=0.5*9.8*t^2
=>200/9.8=t^2

=>t=4.51secs

b) using equation of motion
v=u+a*t

substututing the given values

v=0+9.8*4.51
=>v=9.8*4.51

=>v=44.198 m/s

2)
given data:
intial velocity u=15m/s
acceleration a=-0.30m/s^2 (minus because it is decelerating)
final velocity v=0 m/s (as the train stops)
a) using the equation
v=u+a*t

subtituting the given values
=>0=15+(-0.30)*t
=>t=15/0.30

=>t=50secs

b) s=ut+0.5at^2

s=15*50-0.5*0.3*(50)^2

=>s=750-375
=>s=375m

3) given data:
  
intial velocity u=10m/s
final velocity v=28m/s
distance s=40m

#s=ut+0.5at^2

40=10t+0.5at^2 eq1

#v=u+at

28=10+at
at=18 eq2

rewriting and substituting the value of (at) in eq1 we get
  
40=10t+0.5*(at)*t
=> 40=10t+9t
=>19t=40
=>t=40/19=2.1052

so acceleration a=18/2.1052 or a=8.55m/s^2

4)
given data
u=0;
v=62m/s
a=12m/s^2

v=u+at
=>62=12*t
=>62/12=t;
=>t=5.166secs

s=ut+0.5at^2
s=0*t+0.5*12*(5.166)^2
s=160.16 m

5)
for upward motion
given data
a=-9.8m/s^2
t=2.8s
v=0m/s^2

v=u+at
=>0=u-9.8*2.8
=>u=27.44m/s

h=ut+0.5at^2
=>h=27.44*2.8-0.5*9.8*(2.8)^2

=>h=38.416

for downward motion
given data
a=9.8m/s^2
t=2.8s (as the time taken will be same in both cases)
u=0m/s^2
  
v=u+at
=>v=0+9.8*2.8

=>v=27.44m/s

6)

let the motorist cathes up after t time

distance travelled by police-15=distance travelled by the motorist
25*t-15=38*t-0.5*1*t
=>t^2-26t-30=0
=>t=27.1sec

v=u+at
=>v=38-1*27.1
=>v=10.9m/s

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