Question

Two balls A and B of masses 100 grams and 300 grams respectively are pushed horizontally from a table of height 3 meters. Ball has is pushed so that its initial velocity is 10 m/s and ball B is pushed so that its initial velocity is 15 m/s. 
a) Find the time it takes each ball to hit the ground. 
b) What is the difference in the distance between the points of impact of the two balls on the ground?​

Answer 1

Answer:

Explanation:

hii mate here is ur answer_______

a) The two balls are subject to the same gravitational acceleration and therefor will hit the ground at the same time t found by solving the equation

-3 = -(1/2) g t2

t = √ (3(2)/9.8) = 0.78 s

b) Horizontal distance XA of ball A

XA = 10 m/s * 0.78 s = 7.8 m

Horizontal distance XB of ball B

XB = 15 m/s * 0.78 s = 11.7 m

Difference in distance XA and XB is given by

|XB - XA| = |11.7 - 7.8| = 3.9 m

Answer 2

Given:

Two balls A and B of masses 100 grams and 300 grams respectively are pushed horizontally from a table of the height of 3 meters such that the initial velocity of A is 10 m/s and ball B is  15 m/s.

To Find:

• Time taken by each ball to hit the ground.
• Difference between the distances of points of impact of the two balls.

Solution:

Here, the two balls initially have only the horizontal velocity i.e. the vertical component of velocities of both are equal to zero.

$u_{y_A}=u_{y_B}=0$

Since they experience the same amount of acceleration due to gravity $g=10m/s^2$, they will hit the ground at the same time $t$.

    $y=u_yt+\frac{1}{2} a_yt^2$

⇒ $3=\frac{1}{2} *10*t^2$

⇒ $t=0.78s$

Distance of impact of ball A = Horizontal dIstance traveled by it in time $t$

⇒ $x_A=u_{x_A}t$

⇒ $x_A=10*0.78=7.8m$

Similarly, the distance of impact of ball B,

   $x_B=u_{x_B}t$

⇒ $x_B=15*0.78=11.7m$

The difference in the distance of impact = 3.9m

Hence, both the ball will hit the ground in 0.78s and will have a difference in distance of the impact of 3.9m.