the formula h=15t-5t^2 gives the height 'h' metres of a ball, 't' seconds after it is thrown up into the air.
(i) find the times when the height is 10m
(ii) after how long does the bal hit the ground?
ive figured out that the answer to (i) is t= 1 or 2 by using the quadratic equation, but im still unsure on how to solve (ii)
thanks for your help in advance!
how high is the ball when it hits the ground?
so instead of setting h=10, like you did in (i), set h=0
you should get 5t^2 - 15t = 0
5t^2 - 3t = 0
t(5t-3) = 0
t=0, (at the start) or t=3/5 (when it hits the ground)
<you should get 5t^2 - 15t = 0
5t^2 - 3t = 0
t(5t-3) = 0
t=0, (at the start) or t=3/5 (when it hits the ground) >
should be
you should get 5t^2 - 15t = 0
t^2 - 3t = 0
t(t-3) = 0
t=0, (at the start) or t=3 (when it hits the ground)
If H (the height) = 15t - 5 t^2, then when H = 10,
-5 t^2 +15t^2 -10 = 0
t^2 -3t +2 = 0
(t-1)(t-2) = 0
So your first answer was correct. Thos are the itmes the ball is at 20 m: going up and coming down.
The time when it hits the ground are given by the equation:
-5t^2 +15 t = 0
t(t-3)= 0
t = 0 or 3 seconds.
(t=0 corresponds to when it was thrown)
thanks once again for your help!
maths!
The height in meters of a ball above the ground at any time in second when it is projected is given as h= 15t-t².
At what time will it reach its highest point?
We can find the highest point by finding the vertex of the quadratic equation.
The equation for the height of the ball is h = -t² + 15t.
To find the time at which the ball reaches its highest point, we need to find the t-value of the vertex.
We know that the t-value of the vertex is given by:
t = -b/2a
where a = -1 and b = 15.
Substituting these values:
t = -15/2(-1)
t = 7.5 seconds
Therefore, the ball will reach its highest point at 7.5 seconds after it is projected.