Answer :
Vertically we can say
Vertical acceleration = - g
Vertical velocity = u sinѲ - gt [Ѳ = 42; u = 12 m/s]
Vertical displacement = u sinѲt - (1/2) g t^2 + 9.5
When the rock hits the ground, its vertical displacement will be zero. So we can say..
u sinѲt - (1/2) g t^2 + 9.5 = 0
I'll rearrange this ....
- (1/2) g t^2 + u sinѲt + 9.5 = 0
Can you see that we now have a quadratic in t?
Using the well known formula
t = [ - u sinѲ ± √( u^2 sin^2Ѳ - 4(( - (1/2)g * 9.5))] / (- g)
t = [- u sinѲ ± √( u^2 sin^2Ѳ + 19g)] / (-g)
t = [- 8.03 ± √(250.86)] / (-9.81)
t = [- 8.03 ± 15.84] / (-9.81)
t = (- 8.03 + 15.84) / (-9.81) or t = (-8.03 - 15.84) / (-9.81)
t = - 0.8 or t = 2.43
Well, a negative time has no meaning so
t = 2.43 seconds.
Vertical acceleration = - g
Vertical velocity = u sinѲ - gt [Ѳ = 42; u = 12 m/s]
Vertical displacement = u sinѲt - (1/2) g t^2 + 9.5
When the rock hits the ground, its vertical displacement will be zero. So we can say..
u sinѲt - (1/2) g t^2 + 9.5 = 0
I'll rearrange this ....
- (1/2) g t^2 + u sinѲt + 9.5 = 0
Can you see that we now have a quadratic in t?
Using the well known formula
t = [ - u sinѲ ± √( u^2 sin^2Ѳ - 4(( - (1/2)g * 9.5))] / (- g)
t = [- u sinѲ ± √( u^2 sin^2Ѳ + 19g)] / (-g)
t = [- 8.03 ± √(250.86)] / (-9.81)
t = [- 8.03 ± 15.84] / (-9.81)
t = (- 8.03 + 15.84) / (-9.81) or t = (-8.03 - 15.84) / (-9.81)
t = - 0.8 or t = 2.43
Well, a negative time has no meaning so
t = 2.43 seconds.