Question

a ball projected vertically upward at t is equal to zero the ball is at same height at time 4 second and 8 second the maximum height attained by the ball is​

Answer 1

Answer:

Maximum height = 176.22 m

Explanation:

Given,

• $t_1\ =\ 4 sec$
• $t_2\ =\ 8\ sec$

From the projectile motion, The total time of flight of the object is equal to the sum of the two times where the object is at same height.

Time of flight of the ball,

$t\ =\ t_1\ +\ t_2\\\Rightarrow t\ =\ 4\ +\ 8\ =\ 12\ sec$

At maximum height,

Let u be the initial speed of the ball,

Time taken by the ball to reach the maximum height  = $\dfrac{t}{2}\ =\ 6 sec$

Final speed of the ball = v = 0, and acceleration of the ball = -g

From the kinematics,

$v\ =\ u\ -\ gt\\\Rightarrow 0\ =\ u\ -\ gt\\\Rightarrow u\ =\ gt\ =\ 9.81\times 6\\\Rightarrow u\ =\ 58.9\ m/s.$

Now,

Let h be the maximum height of the ball,

$\therefore h\ =\ ut\ -\ \dfrac{1}{2}gt^2\\\Rightarrow h\ =\ 58.9\times 6\ -\ 0.5\times 9.81\times 6^2\\\Rightarrow h\ =\ 176.22\ m$

Hence the maximum height attained by the ball is 176.22 m