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AS physics/mechanics problems
1) A tennis ball is dropped from a height of 2.0m above a hard level floor and it rebounds to a height of 1.5m. Sketch the v-t graph for the motion of the tennis ball from the instant it is released to the point where it regains maximum height. Calculate;
a) the total time taken
b) the speed just after impact
2)A rocket accelerates from rest at ground level at 7ms^-2 for 30s vertically. Calculate;
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
If anyone can help me with these I will be so grateful.
a) the total time taken
b) the speed just after impact
2)A rocket accelerates from rest at ground level at 7ms^-2 for 30s vertically. Calculate;
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
If anyone can help me with these I will be so grateful.
1) A tennis ball is dropped from a height of 2.0m above a hard level floor and it rebounds to a height of 1.5m. Sketch the v-t graph for the motion of the tennis ball from the instant it is released to the point where it regains maximum height. Calculate;
a) the total time taken
b) the speed just after impact
a) the total time taken
b) the speed just after impact
(a)Calculate the time taken for it to reach the ground with u=0, a=9.8, s=2, t=? using the constant acceleration equations. Add this to the time taken to reach the highest point, using u=?, v=0, a=-9.8, s=1.5, t=?.
(b)Calculate u, the initial velocity with the information above.
2)A rocket accelerates from rest at ground level at 7ms^-2 for 30s vertically. Calculate;
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
If anyone can help me with these I will be so grateful.
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
If anyone can help me with these I will be so grateful.
More constant acceleration questions.
Use the constant acceleration equations. If you don't know them, they are as follows:
v=u+at
v^2=u^2 + 2as
s=ut+0.5a^2
s=0.5(u+v)t
Hope that helps.
Gaz031
(a)Calculate the time taken for it to reach the ground with u=0, a=9.8, s=2, t=? using the constant acceleration equations. Add this to the time taken to reach the highest point, using u=?, v=0, a=-9.8, s=1.5, t=?.
(b)Calculate u, the initial velocity with the information above.
More constant acceleration questions.
Use the constant acceleration equations. If you don't know them, they are as follows:
v=u+at
v^2=u^2 + 2as
s=ut+0.5a^2
s=0.5(u+v)t
Hope that helps.
(b)Calculate u, the initial velocity with the information above.
More constant acceleration questions.
Use the constant acceleration equations. If you don't know them, they are as follows:
v=u+at
v^2=u^2 + 2as
s=ut+0.5a^2
s=0.5(u+v)t
Hope that helps.
that's great, could you do 2)b for me as I really don't understand it.
2)A rocket accelerates from rest at ground level at 7ms^-2 for 30s vertically. Calculate;
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
You need to find the answer to (a) first. Then find the further distance upward it travels when the acceleration is replaced by g.
Add this extra height to the height for a and you have your final answer.
Gaz031
You need to find the answer to (a) first. Then find the further distance upward it travels when the acceleration is replaced by g.
Add this extra height to the height for a and you have your final answer.
Add this extra height to the height for a and you have your final answer.
i'm an idiot. I did that, but forgot to add the other bit on, thanks.
Gaz031
(a)Calculate the time taken for it to reach the ground with u=0, a=9.8, s=2, t=? using the constant acceleration equations. Add this to the time taken to reach the highest point, using u=?, v=0, a=-9.8, s=1.5, t=?.
(b)Calculate u, the initial velocity with the information above.
(b)Calculate u, the initial velocity with the information above.
I've done this but the answer says 6.32ms^-1, but I get a different answer.
Can someone help me?
Question 1:
1a) Time from top to floor:
-s=ut-(1/2)(g)(t^2)
Therefore:
s=-ut+(1/2)(g)(t^2)
u=0 => 2s=g(t^2) => t=root(2s/g)=root((2(2))/9.8)=root(4/9.8)=0.64s (T1)
Greatest height=1.5m, at this v=0
Form two equations:
v=u-gt and v=0 => u=gt (i)
And
s=ut-(1/2)(g)(t^2)
Substituting (i) into (ii):
s=(g)(t)(t)-(1/2)(g)(t^2)=(1/2)(g)(t^2)
Therefore:
t=root(2s/g)=root(2(1.5)/g)=0.55s (T2)
Therefore Ttot=T1+T2=0.64+0.55=1.19s
b) Speed just after the impact is u in (i). Therefore, as we were assuming the time it would take the ball to reach the 1.5m, we substitute T2 into (i):
u=gt=(9.8)(0.55)=5.39m(s^(-1))
This is the DEFINITE answer. The book is wrong.
Newton.
1a) Time from top to floor:
-s=ut-(1/2)(g)(t^2)
Therefore:
s=-ut+(1/2)(g)(t^2)
u=0 => 2s=g(t^2) => t=root(2s/g)=root((2(2))/9.8)=root(4/9.8)=0.64s (T1)
Greatest height=1.5m, at this v=0
Form two equations:
v=u-gt and v=0 => u=gt (i)
And
s=ut-(1/2)(g)(t^2)
Substituting (i) into (ii):
s=(g)(t)(t)-(1/2)(g)(t^2)=(1/2)(g)(t^2)
Therefore:
t=root(2s/g)=root(2(1.5)/g)=0.55s (T2)
Therefore Ttot=T1+T2=0.64+0.55=1.19s
b) Speed just after the impact is u in (i). Therefore, as we were assuming the time it would take the ball to reach the 1.5m, we substitute T2 into (i):
u=gt=(9.8)(0.55)=5.39m(s^(-1))
This is the DEFINITE answer. The book is wrong.
Newton.
The Chameleon
1) A tennis ball is dropped from a height of 2.0m above a hard level floor and it rebounds to a height of 1.5m. Calculate;
a) the total time taken
b) the speed just after impact
a) the total time taken
b) the speed just after impact
a.) On way to ground: s = 2.0, u = 0, a = 9.8.
s = ut + 1/2at^2
---> 2 = 1/2(9.8)t^2 = 4.9t^2
---> t^2 = (2/4.9)
---> t = Sq.Root (2/4.9)
---> t (Reach floor) = 0.639 s
On way to max height of 1.5m: v = 0, s = 1.5, a = -9.8
v^2 = u^2 + 2as
---> 0 = u^2 - 19.6(1.5)
---> 0 = u^2 - 29.4
---> u^2 = 29.4
---> u = 5.42 ms^-1
v = u + at
---> 0 = 5.42 - 9.8t
---> 9.8t = 5.42
---> t = 0.553 s
Hence: Total Time Taken = 0.639 + 0.553 = 1.19 s (3.S.F)
Speed (Just after Impact) = u (on way to max. height) = 5.42 ms^-1 (3.S.F)
cam
2)A rocket accelerates from rest at ground level at 7ms^-2 for 30s vertically. Calculate;
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
a) the speed and height of the rocket at 30s after launch
b) the maximum height gained by the rocket if its fuel is used up after 30s
c) the speed of impact at the ground
d) the time of flight
a.) At 30 s after rocket launch: u = 0, a = 7, t = 30.
v = u + at
---> v = 7 * 30
---> Speed (At 30 s) = 210 ms^-1
s = ut + 1/2at^2
---> s = 1/2(7)(30^2)
---> Height (At 30 s) = 3150 m
b.) From point where fuel is used up:
a = -9.8, u = 210, v = 0.
v^2 = u^2 + 2as
---> 0 = 210^2 - 19.6s
---> s = (210^2)/19.6
---> s = 2250 m
Hence: Max. Height (Rocket) = 3150 + 2250 = 5400 m
c.) From max. height to ground:
u = 0, a = 9.8, s = 5400.
v^2 = u^2 + 2as
---> v^2 = 2(9.8)(5400)
---> v^2 = 105840
---> Speed Of Impact (At Ground) = 325.3 ms^-1 (1.D.P)
d.) From point where fuel is used up:
v = u + at
---> 0 = 210 - 9.8t
---> 9.8t = 210
---> t = 21.43 s
Time (Reach Max. Height) = 30 + 21.43 = 51.43 s
Hence: Time Of Flight = 2 * (Time To Reach Max. Height) = 2 * 51.43
---> Time Of Flight = 102.9 s (1.D.P)
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