Two spheres are launched horizontally from a 1.2\\({rm\\;{\ m m}}\\) -high table

Question

Two spheres are launched horizontally from a 1.2({rm;{ m m}}) -high table. Sphere (A) is launched with an initial speed 4.5 ({ m ;{rm m/s}}). Sphere (B) is launched with an initial speed of 2.5 ({ m ;{ m m/s}}).
What is the time for the sphere (A) to hit the floor?
What is the time for the sphere (B) to hit the floor?
What is the distance that sphere (A) travels from the edge of the table?
What is the distance that sphere (B) travels from the edge of the table? Two spheres are launched horizontally from a 1.2({rm;{ m m}}) -high table. Sphere (A) is launched with an initial speed 4.5 ({ m ;{rm m/s}}). Sphere (B) is launched with an initial speed of 2.5 ({ m ;{ m m/s}}).
What is the time for the sphere (A) to hit the floor?
What is the time for the sphere (B) to hit the floor?
What is the distance that sphere (A) travels from the edge of the table?
What is the distance that sphere (B) travels from the edge of the table?
What is the time for the sphere (A) to hit the floor?
What is the time for the sphere (B) to hit the floor?
What is the distance that sphere (A) travels from the edge of the table?
What is the distance that sphere (B) travels from the edge of the table?

Explanation / Answer

here,
velocity of sphere A, va = 4.5 m/s
velocity of sphere B, vb = 2.5 m/s
height of table, h = 1.2 m

Part A:
from second eqn of motion,
h = va*t + 0.5*g*t^2
1.2 = 4.5*t + 0.5*9.8*t^2

solving for time period to reach ground, t we get
4.9t^2 + 4.5t - 1.2 = 0
t = 0.216 s

Part B
Part A:
from second eqn of motion,
h = va*t + 0.5*g*t^2
1.2 = 2.5*t + 0.5*9.8*t^2

solving for time period to reach ground, t we get
4.9t^2 + 2.5t - 1.2 = 0
t = 0.302 s

Part C:
as horizontal acceleration si zero, ax = 0
x = va*t
x = 4.5*0.216
x = 0.972 m

Part D:
as horizontal acceleration si zero, ax = 0
x = vb*t
x = 2.5*0.216
x = 0.54 m

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