A projectile is thrown horizontally from a tower of 10m to a inclined plane of slope 45°(The slope faces the tower).If it hits the slope at 90° find the speed with which the projectile is thrown?The answer is10(3/2)^1/2
Thankyou!:)
Thankyou!:)
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Are you certain it's 3/2 and not 2/3? That's what I got - here's how.
a = 10m/s^2 (accel. due to gravity)
v = 10*t (vertical velocity at time t)
s = 10 - 5t^2 (height at time t)
So you know that to hit the 45 degree slope at 90 degrees, the vertical speed must be the same as the horizontal speed (the speed at which it was thrown). That gives you the equation
v_initial = 10*t
You also know that the height has to equal the horizontal distance traveled by the projectile because the slope is 45 degrees, i.e. it's like the graph y = x. This tells you that
v_inital*t = 10 - 5t^2
Solve these two equations simultaneously.
v_initial = 10*t can become v_initial*t = 10*t^2
Then you can set the right hand side of the two equations equal to each other:
10*t^2 = 10 - 5t^2
This gets t = (2/3)^1/2
From the earlier equation you know that v_initial = 10*t. Now that you know t just sub it in!
v_inital = 10*(2/3)^1/2
Hope this helps.
a = 10m/s^2 (accel. due to gravity)
v = 10*t (vertical velocity at time t)
s = 10 - 5t^2 (height at time t)
So you know that to hit the 45 degree slope at 90 degrees, the vertical speed must be the same as the horizontal speed (the speed at which it was thrown). That gives you the equation
v_initial = 10*t
You also know that the height has to equal the horizontal distance traveled by the projectile because the slope is 45 degrees, i.e. it's like the graph y = x. This tells you that
v_inital*t = 10 - 5t^2
Solve these two equations simultaneously.
v_initial = 10*t can become v_initial*t = 10*t^2
Then you can set the right hand side of the two equations equal to each other:
10*t^2 = 10 - 5t^2
This gets t = (2/3)^1/2
From the earlier equation you know that v_initial = 10*t. Now that you know t just sub it in!
v_inital = 10*(2/3)^1/2
Hope this helps.