contestada

A projectile is shot horizontally ( x direction) from the edge of a vertical cliff 100 m above level ground. Its initial speed is 75 m/s. Neglecting air resistance. At what angle is the velocity vector measured clockwise from the x axis the instant before it hits the ground

Respuesta :

Answer:

Theta = 30.56degree

Explanation:

  • for vertical component(Vy) of the projectile;
  • given s = 100m, u = 0m/s, Vx = 75m/s

  • from the second equation of motion ; Vy^2 = U^2 + 2as
  • V^y2 = 0 + 2 x 9.81 x 100
  • Vy = square root ( 1962) = 44.29m/s

  • Angle between them theta = arctan ( Vy/vx) = arctan (44.29/75)
  • Theta = arctan ( 0.59059) = 30.56 degree

Answer:

the angle is θ = -0.533 rad

Explanation:

from the equation of vertical motion

vy²=vy₀² + 2*g*y

where

g= gravity =9.8 m/s²

y= height of the cliff = 100 m

then since the projectile has initially no component of velocity in the vertical direction , vy₀=0

then

vy=√(0² + 2*9.8 m/s²* 100 m) = 44.27 m/s

then the angle θ of the velocity vector  will be

tan θ = -vy/vx

θ = tan ⁻¹[(-44.27 m/s)/75 m/s ]= -0.533 rad

then the angle is θ = -0.533 rad

Otras preguntas