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I need someone to help me solve this no 7 n 9
thanks in anticipation

manuel4real:
I need someone to help me solve this
no 7 n 9

thanks in anticipation

and u are too lazy to type the questions out, how do u expect one to see what u posted not even that ur camera is very sharp.

A low level bomber releases a bomb at a height of 50m above the surface of the sea in a horizontal flight at a constant speed of 320km/h.how long does the ball take to fall to the surface?how far(horizontally)of the point of release is the point of impact? what is the angle of sight

A golfer launches a ball in the eastward direction with an initial speed of 30m/s at an upward angle of 34° with the horizontal .what are the components of the instantaneous velocity of the ball in the reference frame of the ground? The x, y, z axes are point east,up and south respectively.

biafraisdead:
and u are too lazy to type the questions out, how do u expect one to see what u posted not even that ur camera is very sharp.

have typed it. can u help me now plz

manuel4real:
A low level bomber releases a bomb at a height of 50m above the surface of the sea in a horizontal flight at a constant speed of 320km/h.how long does the ball take to fall to the surface?how far(horizontally)of the point of release is the point of impact? what is the angle of sight

A golfer launches a ball in the eastward direction with an initial speed of 30m/s at an upward angle of 34° with the horizontal .what are the components of the instantaneous velocity of the ball in the reference frame of the ground? The x, y, z axes are point east,up and south respectively

time it takes the bomb to fall to the surface

considering vertical motion u=0 (initial velocity downward at release)

using H= ut + 0.5gt²

subst. H=50m, g= 10m/s² and u=0


then 50= 0.5×10×t²

t=√10

t= 3.16s


How far is the point of impact

Horizontal Range= U_x×t

since angle of release is zero
U_x=u

horizontal range= 88.89m/s×3.16s

" "= 280.89m

Angle of sight is the same as the angle of depression taking from the horizontal axis
let the angle be x

tanx= 1/u[√(gh/2)]

u=320km/h
in m/s

u=88.89m/s

substitute u, g and h

tanx= 0.1779

x= arctan0.1779


I don't have a scientific calc. here so do it yourself











components of velocity usually refer to horizontal U_x and vertical components U_y


U_x = 30m/s×sin34°


U_y = 30m/s×cos34°




hope this helps