a ball is projected horizontally from a cliff of height 245m and reaches the ground at a horizontal distance of 350m from the foot of the tower. determine the initial velocity V, and the velocity (direction and magnitude) on striking the ground. (take g=10m/s^2)
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projectiles (1 Viewer)
- Thread starter AFGHAN22
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Mountain.Dew
Magician, and Lawyer.
d^2y / dt^2 = -g
dy/dt = -gt + c1. when t=0, dy/dt = 0 (remember it starts at the top of the tower, ball is projected horizontally)
c1 = 0, therefore dy/dt = -gt
y = -1/2gt^2 + c2, when t=0, y = 245, c2 = 245
therefore, y = -1/2gt^2 + 245....(1)
now...for x
d^2x/dt^2 = 0
dx/dt = c3. when t=0, dx/dt = Vi(this is because the particle is projected HORIZONTALLY. You might think of it this way as well...it is Vcos@, but @ = 0. Therefore, Vcos@ = Vi(1) = Vi.) therefore, c3 = 0
x = Vt + c4, when t=0, x = 0. therefore, c4=0
hence, x = Vit....(2)
now..."reaches the ground at a horizontal distance of 350m from the foot of the tower"
when x = 350, y = 0 ==> solve simualtaenous equations for (1) and (2), eliminating t.
(1)...0 = 245 - 1/2gt^2
-245*2 = -gt^2, g = 10
490/10 = t^2, t = 7, since t > 0
SUB t = 7 and x = 350 into (2)...==> 350 = Vi(7)
Vi = 50
note: V = sqrt(dy/dt^2 + dx/dt^2)
now, the ball hits the ground when t = 7.
therefore, using dy/dt = -gt, when t= 7, dy/dt = -70,
dx/dt = Vi=50
therefore, V = sqrt((-70)^2+ 50^2) = 10sqrt(74)m/s upon hitting the ground.
Direction: draw the velocity triangle with dy/dt = -70, dx/dt = 50, and then find &, where & is the angle of direction to the horizontal axis. a diagram is attached.
you will get & = 180 degress - tan^-1 (7/5) to the horizontal.
dy/dt = -gt + c1. when t=0, dy/dt = 0 (remember it starts at the top of the tower, ball is projected horizontally)
c1 = 0, therefore dy/dt = -gt
y = -1/2gt^2 + c2, when t=0, y = 245, c2 = 245
therefore, y = -1/2gt^2 + 245....(1)
now...for x
d^2x/dt^2 = 0
dx/dt = c3. when t=0, dx/dt = Vi(this is because the particle is projected HORIZONTALLY. You might think of it this way as well...it is Vcos@, but @ = 0. Therefore, Vcos@ = Vi(1) = Vi.) therefore, c3 = 0
x = Vt + c4, when t=0, x = 0. therefore, c4=0
hence, x = Vit....(2)
now..."reaches the ground at a horizontal distance of 350m from the foot of the tower"
when x = 350, y = 0 ==> solve simualtaenous equations for (1) and (2), eliminating t.
(1)...0 = 245 - 1/2gt^2
-245*2 = -gt^2, g = 10
490/10 = t^2, t = 7, since t > 0
SUB t = 7 and x = 350 into (2)...==> 350 = Vi(7)
Vi = 50
note: V = sqrt(dy/dt^2 + dx/dt^2)
now, the ball hits the ground when t = 7.
therefore, using dy/dt = -gt, when t= 7, dy/dt = -70,
dx/dt = Vi=50
therefore, V = sqrt((-70)^2+ 50^2) = 10sqrt(74)m/s upon hitting the ground.
Direction: draw the velocity triangle with dy/dt = -70, dx/dt = 50, and then find &, where & is the angle of direction to the horizontal axis. a diagram is attached.
you will get & = 180 degress - tan^-1 (7/5) to the horizontal.