Range = Vo^2*sin(2A)/g = 8^2*sin(60)/10 = 5.54 m. = 1st jump,
Range = 8^2*sin(90)/10 = , 2nd jump.
in his attempt,a long jumper took off from the spring board with a speed of 8ms^-1 at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal.Given that the expression for the horizontal range of the projectile is U^2sin2theta/g, where all the symbols have their usual meaning.show that he gains a distance of 0.8576m in his second attempt (g=10ms^2)?
2 answers
Initial speed up = U sin T
v = U sin T - g t
at top v = 0
t ime up = U sin T/ g
time in air total = 2 U sin T / g
d = Ucos T * 2 U sin T / g
so
d = (2 U^2/g) cos T sin T = (2 U^2/g) * (1/2) sin (2 T)
= (U^2/g)sin (2T) (your notation is impossible to follow)
if U = 8 then U^2/g = 64/10 = 6.4
if T = 30 deg
d = 6.4 sin 60 = 5.54256
if T = 45 deg
d = 6.4 sin 90 = 6.4
difference = .8574
v = U sin T - g t
at top v = 0
t ime up = U sin T/ g
time in air total = 2 U sin T / g
d = Ucos T * 2 U sin T / g
so
d = (2 U^2/g) cos T sin T = (2 U^2/g) * (1/2) sin (2 T)
= (U^2/g)sin (2T) (your notation is impossible to follow)
if U = 8 then U^2/g = 64/10 = 6.4
if T = 30 deg
d = 6.4 sin 60 = 5.54256
if T = 45 deg
d = 6.4 sin 90 = 6.4
difference = .8574
0 answers