A projectile is fired with an initial speed of 12.0 m/s at an angle of 65.0 abov

ID: 1326504 • Letter: A

Question

A projectile is fired with an initial speed of 12.0 m/s at an angle of 65.0 above the horizontal. The object hits the ground 9.50 s later.

a)How much higher or lower is the launch point relative to the point where the projectile hits the ground?

Express a launch point that is lower than the point where the projectile hits the ground as a negative number.

b)To what maximum height above the launch point does the projectile rise?

c)What is the magnitude of the projectile's velocity at the instant it hits the ground?

d)What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Express your answer using two significant figures

here,

initial velocity of projectile , u = 12 m/s

theta = 65 degree

time taken , t = 9.5 s

(a)

let the height of the point be h

h = u * sin(theta) * t - 0.5 * g * t^2

h = 12 * sin(65) * 9.5 - 0.5 * 9.8 * 9.5^2

h = - 338.91 m

the height of the launch point relative to the point where the projectile hits the ground is 338.91 m

(b)

the maximum height , h = (u * sin(theta))^2/2*g

h = (12 * sin(65))^2 /(2 * 9.8)

h = 6.03 m

the maximum height above the launch point is 6.03 m

(c)

let the velocity of projectile at the time before it hits the ground be v

vy = u *sin(theta) - g * t

vy = 12 * sin(65) - 9.8 * 9.5

vy = - 82.22 m/s

horizontal velocity , vx = u * cos(theta)

vx = 5.07 m/s

v = sqrt(5.07^2 + 82.22 ^2)

v = 82.38 m/s

the magnitude of the projectile's velocity at the instant it hits the ground is 82.38 m/s

(d)

theta = arctan(- 82.22/5.07)

theta = - 86.47 degree

the direction of the projectile's velocity at the instant it hits the ground is 86.47 degree clockwise from the +x axis

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