A projectile is launched at an initial angle of 50 o with a speed of v o = 15 m/

Question

A projectile is launched at an initial angle of 50o with a speed of vo = 15 m/s from the top of a hill 5 meters above level ground.

a) Find the x-component and y-components of the initial velocity.

b) Write the equations for the x position and the x velocity as functions of time. Insert numerical values for vox and xo.

c) Write the equations for the y position and the y velocity as functions of time. Insert numerical values for voy and yo.

d) Find the time it takes for the projectile to reach its maximum height. What condition do you impose on the y equations to find this time?

e) What is the projectile’s height above the surface at this time?

f) How long does it take, from the instant of firing, for the projectile to hit the ground? What condition do you impose on the y equations to find this time?

g) Find the range of the projectile from launch to impact.

h) Find the x and y components of the velocity of the projectile as it strikes the ground.

i) What is the magnitude of the velocity of the projectile when it strikes the ground?

j) Make a sketch of the trajectory labeling the times and positions of the projectile at launch, peak, and impact point.

Explanation / Answer

given,

angle = 50 degree

initial velocity = 15 m/s

horizontal component of velocity = v * cos(50)

horizontal component of velocity = 15 * cos(50)

horizontal component of velocity = 9.642 m/s

vertical component = v * sin(50)

vertical vomponent = 15 * sin(50)

vertical component of velocity = 11.49 m/s

for motion in x direction

by second equation of motion

Sx = u * cos(50) * t + 0.5 * a * t^2

since acceleration in horizontal direction is 0

so

Sx = u * cos(50) * t

Sx = 15 * cos(50) * t

also

by third equation of motion

Vx = v * cos (50)

Vx = 15 * cos(50)

for motion in y direction

by second equation of motion

Sy = v * sin(50) * t + 0.5 * g * t^2

Sy = 15 * sin(50) * t + 0.5 * g * t^2

by first equation of motion

Vy = v * sin(50) + g * t

Vy = 15 * sin(50) + g * t

when projectile will reach its maximum height its final velocity will be 0

0 = 15 * sin(50) - g * t

t = 15 * sin(50) / g

t = 1.173 sec

time it takes for the projectile to reach its maximum height = 1.173 sec

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