[填空题]
A cannonball is fired from the top of a tolqhv9f7 ; qok tzv9b4dpf7lyz /2y)+pwer. The rate of ctcx5encg gbo(6w7 .x. hange of the height, h , of the canncbx.o cg. gtwe7n x(56onball above the ground is modelled by
$h^{\prime}(t)=-4 t+20, t \geq 0$,
where h is in metres and t is the time, in seconds, since the moment the cannonball was fired.
1. Determine the time t at which the cannonball reached its maximum height.
After one second, the cannonball is 26 metres above the ground.
2. 1. Find an expression for h(t) , the height of the cannonball above the ground at time t .
2. Hence, determine the maximum height reached by the cannonball.
3. Write down the height of the tower.
4. Calculate the height of the cannonball 4 seconds after it was fired.
The cannonball hits its target on the ground n seconds after it was fired.
5. Find the value of n .≈
6. Determine the total time the cannonball was above the height of the tower.
A second cannonball is fired from exactly halfway up the tower, with the same projectile motion as the first cannonball.
7. Given that both cannonballs land at the same time, determine the length of time between the first cannonball and the second cannonball being fired.
