Projectile launched from 30 m building at 20 m/s and 30°

[BIGEYE]

My son is just starting his advanced highers physics and asks on how best to approach/solve this problem:

A projectile is launched from the top of a building with an initial speed of 20 m s-1 at an angle of 30° to the horizontal. The height of the building is 30 m.
(a) Calculate how long it takes the projectile to reach the ground.
(b) Calculate the velocity of the projectile on impact with the ground, (magnitude and direction).

Appreciate any help.

TIA

 

[xxChrisxx]

This is found by simple constant acceleration kinematic equations (equations of motion). You also allply assumptions, such as no losses due to air resistance etc.

You use the inital conditions to find the time taken to hit the ground from one of the suitable equations.

From this you can find all the other information.

For part b you find the horizontal and vertical components of the velocities (from the equations above). You then use these vectors to build a triagle, and use trigonometry to work out the combined magnitude and angle from horizontal.

 

[BIGEYE]

Thanks, that's got him thinking. To find the time it takes to hit the ground, he will first need to find the height the projectile gets to.

 

[kyiydnlm]

I do not know what you meant by higher physics. However, for your son's best, you need to ask him to try higher mathmatics. It's nearly an one-step question if your son knows how to use higher mathmatics. Trust me, that's the best for your son.

$$\vec{g} = -9.81j, \vec{h}_{0} = 30j, v_{0} = 20, \theta_{0} = 30^o, \vec{v}_{0} = v_{0}cos\theta_{0}i + v_{0}sin\theta_{0}j$$

$$\vec{v} = \int\vec{g}dt = \vec{v}_{0} + \vec{g}t$$

$$\vec{r} = \int\vec{v}dt = \vec{h}_{0} + \vec{v}_{0}t + \frac{1}{2}\vec{g}t^2$$