A projectile is fired from ground level at time t =0 , at an angle with respect

Question

A projectile is fired from ground level at time t=0 , at an angle with respect to the horizontal. It has an initial speed v0 . In this problem we are assuming that the ground is level.

Part A: Find the time tH it takes the projectile to reach its maximum height H .

Part B: Find tR , the time at which the projectile hits the ground after having traveled through a horizontal distance R .

Part C: Find H , the maximum height attained by the projectile.

Part D: Find the total distance R (often called the range) traveled in the x direction; see the figure in the problem introduction.

(Express tH in terms of v0 , , and g (the magnitude of the acceleration due to gravity for parts A-D)

Explanation / Answer

A)
H=v0sint-1/2gt^2
the projectile reaches its maximum height H.when
vsin=v0sin-gt(H)=0 ==>v0sin=gt(H)
so
t(H)=v0sin/g
replacing in the first eq.
H=(v0sin)^2/(2g)

B)
the projectile hits the ground when
t=v0sin/g+v0sin/g
the trajectory is symmetrical respect to the vertical axis crossing the point H
so the total fly time is twice the time to reach the max. altitude
t=2v0sin/g

c)

  H=(Vo^2/2g)(sin THETA)^2

d)

the total distance R (the range) traveled in x direction is
R=v0cost
when
t=2v0sin/g (fly time)
so
R=(2v0cos*v0sin)/g
or
R=v0^2sin(2)/g

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