Saturday

October 25, 2014

October 25, 2014

Posted by **Anonymous** on Thursday, March 13, 2008 at 11:35am.

- Physics -
**drwls**, Thursday, March 13, 2008 at 1:41pmWrite an equation that says that the horozontal distance travelled during the time the ball is above the top of the building equals 60 m. The only unknown will be the launch velocity V.

The time T spent in the air above the top of the building is given by:

V sinA = g T/2

T = 2 V sinA/g

The horizontal distance travelled above the building is

VcosA*T = 2 V^2 sin A cos A /g = 60 m

V^2 sin (2A)/g = 60 m

where A = 37.1 degrees.

Solve for V

V cos A * T' is the distance the ball lands away from the building; T' is the additional time it takes to get to the bottom, 50 m below. Remember that it will start out with a downward velocity component -V sin A when it passes the edge of the building

- Physics -
**tchrwill**, Thursday, March 13, 2008 at 2:07pmA ball is thrown at an angle of 37.1 degrees above the horizontal from a point on the top of a building and 60m from the edge of the building. It just misses the edge of the building and continues on towards the ground. The building is 50m high. At what speed was the ball thrown and how far from the vertical wall of the building does the ball land?

The horizontal distance a projectile travels is derivable from d = Vo^2sin(2µ)/g where d is the horizontal distance in meters(same elevation), Vo is the initial velocity in m/s, g is the acceleration due to gravity = 9.8m/sec.^2 and µ is the elevation of the initial velocity direction in degrees.

Solving for Vo, Vo = sqrt(9.8(60)/sin(74.2º)) = 24.72m/s.

The ball will clear the edge of the building with same speed as it left the launch point and at the same angle , but below the horizontal.

The horizontal component, Vh, of the velocity at that point is 24.72(cos(37.1) = 19.716m/s and the vertical component, Vv, is 24.72(sin(37.1)) = 14.911m/s.

The time to hit the ground from the edge of the building is derivable from h = Vvt + 9.8t^2 where h = 50m, or 50 = 14.911t + 9.8t^2 or 9.8t^2 + 14.911t - 50 = 0. From the quadratic formula, t = 1.622 sec.

The impact distance from the building is therefore d = tVh = 1.622(19.716) = 31.98m.

**Answer this Question**

**Related Questions**

Physics - Hypothetically, you are standing on the balcony of an apartment on the...

Physics - HELP PLEASE. Hypothetically, you are standing on the balcony of an ...

physics - A ball is thrown from the top of a building upward at an angle of 66...

Physics - A ball is thrown from a building standing 30m high, landing 10m away ...

Physics - An objectis thrown upward from the edge of a tall building with a ...

Physics - An object is thrown from the top edge of a building at a speed of 21....

physics - A rock is projected from the edge of the top of a building with an ...

physic - A ball is shot from the top of a building with an initial velocity of ...

physics134 - A ball, with a mass of 0.144 kg, is thrown from the top of a 76.8 m...

physics - A ball is thrown horizontally with an initial velocity of 20.0 m/s ...