Asked by dave
The height h, in metres, of a ball thrown from a cliff can be modelled by the function with the rule h(t) = 20+15t-5t^2, where t less than/equal to 0 is the time measured in seconds.
a) find the initial height of the ball
b) find the time at which the ball hits the ground
Thanks would be highly appreciated!
a) find the initial height of the ball
b) find the time at which the ball hits the ground
Thanks would be highly appreciated!
Answers
Answered by
Reiny
initial ----> t = 0
So what do you get when you replace t with 0 in
h(t) = 20 + 15t - 5t^2 ?
When it hits the ground, isn't the height zero?
0 = 20 + 15t - 5t^2
divide by -5
t^2 - 3t - 4 = 0
solve the quadratic using whatever method you know, make sure to reject
the negative value of t
( it factors, you should be able to see the factors in your head)
So what do you get when you replace t with 0 in
h(t) = 20 + 15t - 5t^2 ?
When it hits the ground, isn't the height zero?
0 = 20 + 15t - 5t^2
divide by -5
t^2 - 3t - 4 = 0
solve the quadratic using whatever method you know, make sure to reject
the negative value of t
( it factors, you should be able to see the factors in your head)
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