Motion in two dimention problem

[Ammar w]

Homework Statement

A projectile is fired from the top of a cliff of height h above the ocean below. The projectile is fired at an angle θ above the horizontal and with an initial speed vi. (a) Find a symbolic expression in terms of the variables vi, g, and θ for the time at which the projectile reaches its maximum height. (b) Using the result of part (a), find an expression for the maximum height hmax above the ocean attained by the projectile in terms of h, vi, g, and θ.

Homework Equations

v_fy = v_iy - gt = v_isinθ_i - gt

the maximum hight = \frac{vi^2 (sinθi)^2}{2g}

The Attempt at a Solution

a)
v_fy = v_iy - gt = v_isinθ_i - gt

it reaches the maximum hight when v_fy = 0.

=> v_isinθ_i - gt = 0

t = \frac{ visinθi}{g}

right??

b) h_max = v_iyt - \frac{1}{2}gt²

h_max = v_isinθ_i (\frac{ visinθi}{g}) - \frac{1}{2}g (\frac{ visinθi}{g})²

h_max = ... how to simplify it??

 

[tiny-tim]

Hi Ammar w!

(btw, there's no need to write θ_i …

there's only one θ, so just write θ ! )

h_max = v_isinθ_i (\frac{ visinθi}{g}) - \frac{1}{2}g (\frac{ visinθi}{g})²

h_max = ... how to simplify it??

just expand that second bracket!

(and then get some sleep! :zzz:)

btw, if you know the standard constant acceleration equations, you should be able to find one that solves the problem straight away​

 

[SammyS]

Homework Statement

A projectile is fired from the top of a cliff of height h above the ocean below. The projectile is fired at an angle θ above the horizontal and with an initial speed vi. (a) Find a symbolic expression in terms of the variables vi, g, and θ for the time at which the projectile reaches its maximum height. (b) Using the result of part (a), find an expression for the maximum height hmax above the ocean attained by the projectile in terms of h, vi, g, and θ.

Homework Equations

v_fy = v_iy - gt = v_isinθ_i - gt

the maximum height = \frac{vi^2 (sinθi)^2}{2g}

The Attempt at a Solution

a)
v_fy = v_iy - gt = v_isinθ_i - gt

it reaches the maximum hight when v_fy = 0.

=> v_isinθ_i - gt = 0

t = \frac{ visinθi}{g}

right??

b) h_max = v_iyt - \frac{1}{2}gt²

h_max = v_isinθ_i (\frac{ visinθi}{g}) - \frac{1}{2}g (\frac{ visinθi}{g})²

h_max = ... how to simplify it??

Your solution to part a) looks fine.

For part b, what is the height (above the ocean) of the projectile at time t = 0 ? It's h, correct?

So you need to modify the equation, h_max = v_iyt - \frac{1}{2}gt² to reflect that.