i have been working on this problem but i just cant seem to get the write answer. could someone work it out and show me the working so i know where i am going wrong.
A ball is thrown verticallu downwards with speed 5ms^-1 from the top of a tower block 46m above the ground. at th same time as A is throw downwards, another ball B is thrown vertically upwards from the ground with speed 18ms^-1. the balls collide. Find the distance of the point where A and B collide from the point where A was thrown.
what i did was s=ut+0.5at^2 for each of them set them equil worked out t using the quadratic formula and put t into the s=ut+0.5at^-2 to get s
i them did 46-s
but my s was to large.
the actual answer is 30m
please show me how to get to this. write out all the formula you use before you use them.
A ball is thrown verticallu downwards with speed 5ms^-1 from the top of a tower block 46m above the ground. at th same time as A is throw downwards, another ball B is thrown vertically upwards from the ground with speed 18ms^-1. the balls collide. Find the distance of the point where A and B collide from the point where A was thrown.
what i did was s=ut+0.5at^2 for each of them set them equil worked out t using the quadratic formula and put t into the s=ut+0.5at^-2 to get s
i them did 46-s
but my s was to large.
the actual answer is 30m
please show me how to get to this. write out all the formula you use before you use them.
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When the balls meet, the sum of their distances is 46.
Let A be the distance of the downward ball and B be the upward ball, then
A= 5t +0.5gt² {g is the acceleration due to gravity, approx. 9.8m/s²}
B=18t - 0.5gt² {going against gravity, so the acceleration is negative},
they will meet, after t seconds, when A + B= 46, i.e.
5t +0.5gt² + 18t - 0.5gt² = 46, so
t=2 seconds {collision time}
Now put this value into A to get the distance A drops.
A=5(2) +0.5g(2)²
= 10 + 2g
= 10 + 19.6
= 29.6 {sometimes g is approximated to 10 which will give you 30).
Let A be the distance of the downward ball and B be the upward ball, then
A= 5t +0.5gt² {g is the acceleration due to gravity, approx. 9.8m/s²}
B=18t - 0.5gt² {going against gravity, so the acceleration is negative},
they will meet, after t seconds, when A + B= 46, i.e.
5t +0.5gt² + 18t - 0.5gt² = 46, so
t=2 seconds {collision time}
Now put this value into A to get the distance A drops.
A=5(2) +0.5g(2)²
= 10 + 2g
= 10 + 19.6
= 29.6 {sometimes g is approximated to 10 which will give you 30).
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let us take the downward direction as positive, measure distances from A, and take g as 10m/s²
s = 5t + 5t²
s = 46 - (18t - 5t²)
equating the two,
23t = 46, t = 2s
s = 5*2 + 5*2² = 30m <------- qed
s = 5t + 5t²
s = 46 - (18t - 5t²)
equating the two,
23t = 46, t = 2s
s = 5*2 + 5*2² = 30m <------- qed
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For going down s1=5t+5t^2 since the acceleration is 10ms^-2,
Going up s2=18t-5t^2 since acceleration is -10ms^-2
s1+s2=46 giving 23t=46 and t=2.and you want s1=10+20=30 as hoped.
Going up s2=18t-5t^2 since acceleration is -10ms^-2
s1+s2=46 giving 23t=46 and t=2.and you want s1=10+20=30 as hoped.