Write 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
A ball is thrown at an angle of 37.1degrees 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?
2 answers
A 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.
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.
1 answer