Sam throws a ball to John. At the Same time John has jumped off (with initial velocity = 0 m/s) a 10m high diving board. Sam is standing by the side of the pool 10m away from John. If Sam throws the ball with 10m/s and John catches it as soon as he hits the water in the pool, what is the maximum height the ball reaches.
If John is at x = 0m then Sam is at x = -10m
Anyway, I was using
dx = vt + 0.5at^2 (v is the initial velocity)
now the initial velocity is given but with no angle of trajectory so it is not possible to find the vertical component of the velocity. The answer key states that dx = 0.5at^2 so the answer is 2.5 but they'e ignored the initial vertical velocity! Help me find where I am going wrong here, please.
If John is at x = 0m then Sam is at x = -10m
Anyway, I was using
dx = vt + 0.5at^2 (v is the initial velocity)
now the initial velocity is given but with no angle of trajectory so it is not possible to find the vertical component of the velocity. The answer key states that dx = 0.5at^2 so the answer is 2.5 but they'e ignored the initial vertical velocity! Help me find where I am going wrong here, please.
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the key state dx=0.5at^2 is the the initial velocity of JoHN is zero not sam,,, so u have the dx and a.....so u can work out the time at which john need to reach the pool,
the ball and john reached the pool at the same time, so the time needed to reach the maximum height for the ball is half the total time, and therefore using ( for sam) : a= (v-u)/t, where v=0 because its at the top of motion so instantaneous rest, u can work out the initial vertical speed,,,,then use v^2 = u^2 + 2as, also for vertical sam ball, v=0, and u already found u and a is 9.8,, all u have left is the maximum displacement...
if u had any troubles doing the calculations just put it in the additional information and ill check that
the ball and john reached the pool at the same time, so the time needed to reach the maximum height for the ball is half the total time, and therefore using ( for sam) : a= (v-u)/t, where v=0 because its at the top of motion so instantaneous rest, u can work out the initial vertical speed,,,,then use v^2 = u^2 + 2as, also for vertical sam ball, v=0, and u already found u and a is 9.8,, all u have left is the maximum displacement...
if u had any troubles doing the calculations just put it in the additional information and ill check that
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wlcm!
2 things to comment on
1)ur 1st pt is true but its v-u NOT average velocity which is (v-u)/2
2)h/2 b y john does not equal to max h by ball, bcoz initial velocity differ, and direction of john didnt change,so speed of john will be greater after some time. calculate it to see the diffrnt
2 things to comment on
1)ur 1st pt is true but its v-u NOT average velocity which is (v-u)/2
2)h/2 b y john does not equal to max h by ball, bcoz initial velocity differ, and direction of john didnt change,so speed of john will be greater after some time. calculate it to see the diffrnt
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Correction added - I put the wrong units for the max height!
Find how long John takes to hit the water. Using h for vertical distance gives:
h = ½gt² (because John's initial velocity is zero).
t = √(2h/g)
= √(2x10/g)
= √(20/g)
= 2√(5/g)
During the time John is in the air, the ball will reach max height and come down again. So we know the time taken for the ball to drop from max height, H, is half of 2√(5/g), i.e √(5/g) seconds.
Then just find how far something falls in time √(5/g). (Because initial vertical velocity is zero when starting from the highest point).
H=½gt²
= ½ x g x [√(5/g)]²
= ½ x g x 5/g
= 2.5m (note g cancels and the ball's initial speed is not needed)
Find how long John takes to hit the water. Using h for vertical distance gives:
h = ½gt² (because John's initial velocity is zero).
t = √(2h/g)
= √(2x10/g)
= √(20/g)
= 2√(5/g)
During the time John is in the air, the ball will reach max height and come down again. So we know the time taken for the ball to drop from max height, H, is half of 2√(5/g), i.e √(5/g) seconds.
Then just find how far something falls in time √(5/g). (Because initial vertical velocity is zero when starting from the highest point).
H=½gt²
= ½ x g x [√(5/g)]²
= ½ x g x 5/g
= 2.5m (note g cancels and the ball's initial speed is not needed)