help please?

Question

the height of a ball as it is thrown upwards at 25m/s is given by the equation:
H = 2 - 2t62 + 25t
where H= the height of the ball above the ground in metres
and t= time in seconds

Using only algebraic methods:
a) find the starting height the ball is thrown upwards from.
b) find the time it takes for the ball to reach it's maximum height and hence, find the greatest height reached.
c) how long until the ball hits the ground?
d) at what times will the ball be 15metres above the ground

p.s please do full working out

thank you

Help please?
the height of a ball as it is thrown upwards at 25m/s is given by the equation:
H = 2 - 2t^2 + 25t
where H= the height of the ball above the ground in metres
and t= time in seconds

Using only algebraic methods:
a) find the starting height the ball is thrown upwards from.
b) find the time it takes for the ball to reach it's maximum height and hence, find the greatest height reached.
c) how long until the ball hits the ground?
d) at what times will the ball be 15metres above the ground

p.s please do full working out

Answer 1

a) substitute t=0 into the equation and get the answer h=2 metres
b) at the maximum dh/dt=0. dh/dh = 25-4t. t=25/4 seconds
substituting t=25/4 into the equation will give you the maximum height
c) likewise put h=0 and solve for t
d) likewise put h=15 and solve for t.

Answer 2

a) The starting height is the height at time 0.

H=2 -2t^2 +25t
@t = 0 H= 2 - 0 + 0 = 2

b) to find the maximum set the first derivative equal to zero and solve for t.

dH/dt =-4t +25
0 = -4t +25
4t = 25
t= 6.25 seconds to reach maximum height

to find maximum height put the time to reach maximum height into the original equation and solve for H:

H=2-2t^2 +25t
@ t = 6.25 H = 1 -2(6.25)^2 +25(6.25)
H= 1 - 78.125 + 156.25
H= 79.125 m is the greates height reached.

c) when it hits the ground H=0 so set the expression for H = 0 and solve for t:

H = 2 - 2t^2 +25t
@H=0 0=2-2t^2 +25
put into standard form: 2t^2 -25t -2 =0
solve with quaratic formula:

t = (25 +/- sqrt(25^2 - 4(2)(-2)))/2(2)
t = (25 +/- sqrt(625 + 16))/4
t= (25 +/- 25.318) /4
ignore the negative root because negative time is meaningless:
t = 50.318/4 = 12.48 seconds (approx) to hit ground

d) set H = 15 and solve for two roots:
15 = 2 -2t^2 +25t
2t^2 -25t +13 = 0
t =(25 +/-)sqrt(25^2-4(2)(13))/2(2)
t= (25 +/-(sqrt(625-104))/4
t =(25 +/-(sqrt 521))/4
t =(25 +/- 22.825)/4
t= 47.825/4 and 2.175/4
t = 11.956 and t = .544 these are the times at which the ball will be 15 m above ground.

The steps are correct but you should check through the work for arithmatic errors.

Answer 3

H = 2 - 2t^2 + 25t
a)This will be at t=0
=>H=2-0+0=2m

b)We make a quadratic and check for maximum value
H = 2 - 2t^2 + 25t
=>H=-2(t^2 - (25/2)t - 1)
=>H=-2(t^2 - (25/2)t - 1 + 625/16 - 625/16)
=>H=-2[(t-25/4)^2 - (641/16)]

=>H=641/8 - 2(t-25/4)^2
This will be maximum if t=25/4 sec

=>H(max)=641/8 metres

c)T=Time taken to reach maximum ht. + time taken to come down
=25/4 + 25/4=12.5 sec

d)15=2-2t^2 + 25t
=>2t^2 - 25t + 13=0
=>t=[(25) (+-)sqrt(625 - 104)]/4
=>t=[25(+-)22.83]/4
=>t=2.17/4=0.542 sec

Answer 4

a) at t=0, h=2 meters
b) at max height, acceleration =0
h=2+25t
c) h= 0= 2-2t^2 +25t
d) 15=2-2t^2+25t

Answer 5

a) at the start t=0
h=2m
b) max height when v=0
v=dH/dt
v= -4t+25
put v=0
0= -4t+25
t= 25/4 s
H= 2-2(25/4)^2+25(25/4)=80.125m
d)put H=15