A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal.

A Projectile Is Shot From The Edge Of A Cliff

Determine the following quantities:

– The horizontal and vertical components of the velocity vector.

– The maximum height reached by the projectile above the launch point.

The aim of this question is to understand the different parameters during 2D projectile motion.

The most important parameters during the flight of a projectile are its range, time of flight, and maximum height.

The range of a projectile is given by the following formula:

R = vi2 sin(2θ)g

The time of flight of a projectile is given by the following formula:

t = 2vi sinθg

The maximum height of a projectile is given by the following formula:

h = vi2 sin2θ2g

The same problem can be solved with the fundamental equations of motion. Which are given below:

vf = vi+at

S=vit+12at2

vf2 = vi2+2aS

Expert Answer

Given that:

vi = 65 m/s

θ = 37

hi = 125 m

Part (a) – The horizontal and vertical components of the velocity vector.

vix = vicos(θ) = 65cos(37) = 52 m/s

viy = visin(θ) = 65sin(37) = 39 m/s

Part (b) – The maximum height reached by the projectile above the launch point.

For upward motion:

vi = 39 m/s

vf = 0 m/s

a = 9.8 m/s2

Using 3rd equation of motion:

S = vf2vi22a

S = 023922(9.8)

S = 152119.6

S = 77.60 m

Numerical Result

Part (a) – The horizontal and vertical components of the velocity vector:

vix = 52 m/s

viy = 39 m/s

Part (b) – The maximum height reached by the projectile above the launch point:

S = 77.60 m

Example

For the same projectile given in the question above, find the time elapsed before hitting the ground level.

For upward motion:

vi = 39 m/s

vf = 0 m/s

a = 9.8 m/s2

Using 1st equation of motion:

t1 = vfvia

t1 = 0399.8

t1 = 3.98 s

For downward motion:

vi = 0 m/s

S = 77.60+125 = 180.6 m

a = 9.8 m/s2

Using 2nd equation of motion:

S = vit2+12at22

180.6 = (0)t2+12(9.8)t22

180.6 = 12(9.8)t22

t22 = 36.86

t2 = 6.07 s

So the total time:

t = t1+t2 = 3.98+6.07 = 10.05 s

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