Thread: [SOLVED] Problem with differenciation

Hi! Please help me with this problem!

A stone is thrown upwards. It is h metres high after t seconds.
h=20t-5t2

So far I found:
a)When is the stone 15 metres high?

15= 20t-5t2
0=4t-t2-3
(t-3)(t-1)

So t=1 and 3

b)When is it highest?

c)Distance travelled(total)?

Using first derivative test, since we need maximum height(say 'h') in time 't',
dh/dt = 0
d(20t - 5t^2)/dt = 0
20 - 5*2t = 0 [Using constant rule ; dt/dt = 1 ; power rule]
10t = 20
t = 2
Therefore, for t = 2 seconds, height is maximum

Putting t = 2 in equation for h,

h = (20*2 - 5 *(2)^2) m
= (40 - 5*4) m
= 20 m

For total distance, since stone reaches a maximum distance of 20 m above ground, total distance is (20 + 20) m, i.e. 40 m.(since stone reaches highest point, then falls back to the ground.)

Originally Posted by thegirlwhoneedshelp

Hi! Please help me with this problem!

A stone is thrown upwards. It is h metres high after t seconds.
h=20t-5t2

So far I found:
a)When is the stone 15 metres high?

15= 20t-5t2
0=4t-t2-3
(t-3)(t-1)

So t=1 and 3

b)When is it highest?

c)Distance travelled(total)?

Hi,

Highest distance is when dh/dt =0

So u will get 20-10t=0 ...solve for t.

Second, find h for that t, means it took t seconds to go upward, then covering the same distance back to its initial position

Hope it helps

Hi! I don't understand how you got the answer in blue below and what the constant rule and the power rule are or have to do with it. Could you explain it for me please?

Originally Posted by fardeen_gen

Using first derivative test, since we need maximum height(say 'h') in time 't',
dh/dt = 0
d(20t - 5t^2)/dt = 0
20 - 5*2t = 0 [Using constant rule ; dt/dt = 1 ; power rule]
10t = 20
t = 2
Therefore, for t = 2 seconds, height is maximum

Putting t = 2 in equation for h,

h = (20*2 - 5 *(2)^2) m
= (40 - 5*4) m
= 20 m

For total distance, since stone reaches a maximum distance of 20 m above ground, total distance is (20 + 20) m, i.e. 40 m.(since stone reaches highest point, then falls back to the ground.)

Hello,

Originally Posted by thegirlwhoneedshelp

Hi! I don't understand how you got the answer in blue below and what the constant rule and the power rule are or have to do with it. Could you explain it for me please?

Constant rule : if a is a constant (here, 5), then you have

Power rule : (here, n=2)

Hi! when I wrote 5t2, I meant 5t squared. Could you tell me what to do then?

Originally Posted by Kai

Hi,

Highest distance is when dh/dt =0

So u will get 20-10t=0 ...solve for t.

Second, find h for that t, means it took t seconds to go upward, then covering the same distance back to its initial position

Hope it helps

Hi! when I wrote 5t2, I meant 5t squared. Could you tell me what to do then?

Originally Posted by thegirlwhoneedshelp
Hi! Please help me with this problem!

A stone is thrown upwards. It is h metres high after t seconds.
h=20t-5t2

So far I found:
a)When is the stone 15 metres high?

15= 20t-5t2
0=4t-t2-3
(t-3)(t-1)

So t=1 and 3

If you meant 5t squared, then how did you get the above equation?(Correct me if I am wrong, but by 5t squared, you do mean (5t) * (5t), don't you?)