a particle is projected at angle tita to the horizontal with initial velocity u.if the horizontal distance and maximum height reached are 20m and 10m respectively. find the value of tita and u
2 answers
H=u^2÷g
Y^2 = Yo^2 + 2g*h = 0
Yo^2 + (-19.6)10 = 0
Yo = 14 m/s = vertical component of initial velocity.
Y = Yo + g*Tr = 0
14 + (-9.8)Tr = 0
Tr = 1.43 s. = Rise time.
Tf = Tr = 1.43 s. = Fall time.
T = Tr+Tf = 1.43+1.43 = 2.86 s. = Time in air.
d = Xo*T = 20
2.86Xo = 20
Xo = 7.0 m/s = horizontal component of initial velocity.
a. Vo = Xo + Yoi = 7 + 14i = 15.7 m/s[63.4o] = Initial velocity.
b. Theta = 63.4o.
Yo^2 + (-19.6)10 = 0
Yo = 14 m/s = vertical component of initial velocity.
Y = Yo + g*Tr = 0
14 + (-9.8)Tr = 0
Tr = 1.43 s. = Rise time.
Tf = Tr = 1.43 s. = Fall time.
T = Tr+Tf = 1.43+1.43 = 2.86 s. = Time in air.
d = Xo*T = 20
2.86Xo = 20
Xo = 7.0 m/s = horizontal component of initial velocity.
a. Vo = Xo + Yoi = 7 + 14i = 15.7 m/s[63.4o] = Initial velocity.
b. Theta = 63.4o.
3 answers