Question

A projectile is launched with a speed of 45 m/s at an angle of 35 degrees above the horizontal from the top of a wall. The projectile lands 250m from the base of the wall.

a.) Determine the height of the wall.

b.) Determine the impact velocity of the projectile

Answer 1

a.

given, initial speed = u = 45 m/s

angle of projection = A = 35 deg

Now, using second kinematics law in horizontal direction,

Sx = ux*t + 0.5*ax*t^2

here, t = time taken to hit the ground = ??

ux = horizontal component of initial speed = u*cosA = 45*cos(35 deg)

ax = horizontal acceleration = 0 m/s^2

Sx = horizontal displacement = 250 m

then, 250 = 45*cos(35 deg)*t

t = 250/(45*cos(35 deg))

t = 6.782 s

By using second kinematics law in vertical direction,

Sy = uy*t + 0.5*ay*t^2

here, t = time taken to hit the ground = 6.782 s

uy = vertcial component of initial speed = u*sinA = 45*sin(35 deg)

ay = vertical acceleration = -9.81 m/s^2

Sy = vertical displacement = -height of wall = ?? m

So, Sy = 45*sin(35 deg)*6.782 - 0.5*9.81*6.782^2 = -50.558 m

then, heigt of wall = -Sy = 50.56 m

b.

Let, final velocity = v at angle cw from horizontal

Since, there is zero acceleration in horizontal direction.

So, horizontal component of speed remain constant.

then, vx = ux = 45*cos(35 deg) = 36.86 m/s

now, by using first kinematics law in vertical direction,

vy = uy + ay*t

vy = 45*sin(35 deg) - 9.81*6.782 = -40.72 m/s

So, final velocity = v = sqrt(vx^2 + vy^2) = sqrt(36.86^2 + 40.72^2) = 55 m/s

direction = = arctan(40.72/36.86) = 48 deg

therefore, impact velocity of the projectile = 55 m/s at 48 degrees below the horizontal

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