science help (1 Viewer)

[d3st1nyLiang]

I don't quite get how to do these questions, thanks for any help

1. A ball is thrown vertically up in the air at an initial speed of 50m/s. Calculate

a) maximum height b) displacement at 3.5s c) time of flight

2. Shane Warne bowls a ball at 42m/s towards Sachin Tendulkar, who is then able to hit in the opposite direction at a speed of 50m/s. Calculate the change in momentum of the ball, if its mass is 500grams

hence, calculate the average force provided by the bat, if it is in contact with the ball for 0.1seconds

thanks

 

[bored.of.u]

I don't quite get how to do these questions, thanks for any help

1. A ball is thrown vertically up in the air at an initial speed of 50m/s. Calculate

a) maximum height b) displacement at 3.5s c) time of flight

2. Shane Warne bowls a ball at 42m/s towards Sachin Tendulkar, who is then able to hit in the opposite direction at a speed of 50m/s. Calculate the change in momentum of the ball, if its mass is 500grams

hence, calculate the average force provided by the bat, if it is in contact with the ball for 0.1seconds

thanks

For 2:
change in momentum = m(v-u)
change in momentum = 0.5(42-(-50))
change in momentum = 46 Ns

Impulse = change in momentum = Ft
46 = F(0.1)
F = 460 N

 

[d3st1nyLiang]

ohk thanks

 

[aya-chan]

I don't quite get how to do these questions, thanks for any help

1. A ball is thrown vertically up in the air at an initial speed of 50m/s. Calculate

a) maximum height b) displacement at 3.5s c) time of flight

thanks

a) max height = displacement/2
where v = 0
u = 50 m/s
aand ... uhm. oh yeah. a = -9.8

and then sub that into ... the formula which I'm assuming you know.
I feel especially lazy today.

 

[aya-chan]

v^2 = u^2 + 2as ... if you were wondering.

so ... 0 = 50^2 + 2 x -9.8 s

s = 127.551 ...

so max height = 63.776 ...

=D

well anyway, b) t = 3.5s ...
so use ...:
s = ut + ½ at

and you should get ...
s = 157.85

and then ... c) use v = u + at
where v = 0
u = 50m/s
a = -9.8

so ... t = 5.10...s

 

[d3st1nyLiang]

ohhk, i did something weird , thanks for the help

 

[d3st1nyLiang]

For 2:
change in momentum = m(v-u)
change in momentum = 0.5(42-(-50))
change in momentum = 46 Ns

Impulse = change in momentum = Ft
46 = F(0.1)
F = 460 N

why is u = -50? and v = 42?

 

[aya-chan]

coz they're in opposite directions.

 

[d3st1nyLiang]

i mean why is the initial not 50?

 

[aya-chan]

well, you could make the initial 50 but then the other would be -42. It really just depends which direction you decide to be positive.

oh and you have to put the direction with it too. *forgot* so like
1a) 63.776 m up.
b) hey wait ... I just noticed... I must've done something wrong with the calculations somewhere.
c) is fine.

and 2. would be 460N towards Shane Warne

 

[bored.of.u]

i mean why is the initial not 50?

it doesnt matter which direction you define to be positive they both produce the same answer. I just defined 42 to be positive.

 

[d3st1nyLiang]

so how do u do 1 b) ??

 

[thisnameistaken]

so how do u do 1 b) ??

Mr. Liang

y = -gt^2 / 2 + vtsin(theter) < through maths projectile motion

or

r = ut + 1/2(a)(t^2) < through a physics formula





whats the answer??

 

[aya-chan]

There we go. that's what I left out. I left out the squared on the last t.
whoops.