Question

A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reaches above its starting point in terms of v0, θ0, and g. (Ignore any effects due to air resistance.)

Answer 1

Answer:

$h= \frac{(v_{o})^{2} sin^{2} \theta o }{2g}$

Explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

vfy²= v₀y²-2gH Equation (4)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀  : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y²$h= \frac{(v_{o})^{2}sin^{2} \theta o }{2g}$

2gh=  (v₀*sin θ₀)²

$h= \frac{v_{o}^{2} sin^{2} \theta o }{2g}$