An object is launched from the ground with an initial velocity of v m/s at an an

Question

An object is launched from the ground with an initial velocity of v m/s at an angle of of above the horizontal. t seconds later the object lands on the top edge of a cliff (marked with a) h meters above the ground.

1. what is the height h of the cliff?
2. what is the velocity of the object just as it reaches the top of the cliff?
write the velocity in i, j componets from.
Vf=....i+....j

3. what is the speed of the object just as it reaches the top of the cliff?
4. with what angle (relative to the horizontal) does the object impact the top of the cliff? (the angle of impact is determined by velocity vector at impact.)
5. write the parametric equations of motion x(t)= and y(t)=

[Vo=62, =35, t=5.1]

An object is launched from the ground with an initial velocity of v m/s at an angle of of θ above the horizontal. t seconds later the object lands on the top edge of a cliff (marked with a) h meters above the ground. what is the height h of the cliff? what is the velocity of the object just as it reaches the top of the cliff? write the velocity in i, j components from. Vf=....i+....j what is the speed of the object just as it reaches the top of the cliff? with what angle (relative to the horizontal) does the object impact the top of the cliff? (the angle of impact is determined by velocity vector at impact.) write the parametric equations of motion x(t)= and y(t)=

Explanation / Answer

Given:

Initial speed Vo = 62 m/s

Angle theta = 35 degrees

Time t = 5.1 s

a)

Vertical component of velocity Vo_y = 62*sin35 = 35.56 m/s

Horizontal component of velocity Vo_x = 62*cos35 = 50.79 m/s

After 5.1 s, horizontal distance travelled = 50.79*5.1 = 259 m

It will reach ma.x height when V_y = 0.

Using v = u + at for vertical direction,

0 = 35.56 - 9.81*t

t = 3.62 s

Remaining time = 5.1 - 3.62 = 1.48 s.

Max. height is given by H_max = 35.56^2 / (2*9.81) = 64.45 m

From max. height, Vertical distance dropped in 1.48 s is given by H = ut + 1/2*gt^2

H = 0 + 1/2*9.81*1.48^2 = 10.744 m

Hence, height of cliff h = 64.45 - 10.744 = 53.7 m

b)

Vertical velocity at time of hitting = 9.81*1.48 = 14.52 m/s

Hence, V = 50.79 i - 14.52 j

c)

Speed = sqrt (50.79^2 + 14.52^2) = 52.82 m/s

d)

angle of impact = atan (-14.52 / 50.79) = -15.95 deg

e)

x(t) = 50.79*t

y(t) = 35.56*t - 1/2*g*t^2

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