Question

If the initial velocity stays the same, but the height the projectile is launched from changes , how would the amount of time change that it takes for the projectile to reach its maximum height? How would its maximum height change?

Answer 1

Answer:

Part A

The height to which the projectile reaches is given by the following formula;

$\Delta y = y-y_0 = v_{0y} \times t-\dfrac{1}{2} \times g \times t^2$

Where;

$v_{0y}$ = The initial velocity of the projectile = Stays the same

y = The height the projectile reaches

y₀ = The height from which the projectile is launched

g = The acceleration due to gravity

t = The time taken

Given that the projectile has the same initial velocity, the variation of Δy with time, 't', will be the same when the project is launched from a different height, and the time change that it takes for the projectile to reach the maximum height will be the same for the previous launch height

Part B

From the change in eight of the projectile, Δy = y - y₀, the maximum height reached, 'y', is therefore given as follows;

y = Δy + y₀

The maximum height will increase by 'y₀', where 'y₀' is the difference between new height and the previous height, $y_i$

That is, if y₀ > $y_i$, the maximum height reached increases and if y₀ < $y_i$, y₀ will be added as y = Δy - y₀ and the maximum height reached decreases

Step-by-step explanation: