Question

a ball is thrown vertically upward with a velocity of 19.6 m/s.find the maximum height reached by the ball

Answer 1

Given:velocity =19.6m/s
 Calculating the time taken to reach the highest point:
v = u + at 
0 = 19.6 - 9.8 t 
t = 2 sec
Distance travelled:
s = ut + at²/2 = 19.6 ×2 + 9.8 × 2²/2  
s = 19.6×2 + 19.6 = 19.6×3 = 58.8 m 

Answer 2

Answer:

A ball is thrown in a vertically upwards direction with a velocity of 50 m/s. Maximum height covered  by the ball will be $19.208\ m$

Explanation:

We can solve this problem by using the third law of motion.

• According to the third law of motion,

        $v^{2} - u^{2} =2as$    ...(1)

• where v is the final velocity of the object, u is the initial velocity of the object, a is the acceleration and s is the distance covered by the object.

In the given problem,

$s = h_{max}$  

$a =- g$

(since the ball is thrown upward)

where g is the acceleration due to gravity and $h_{max}$ is the maximum height covered by the ball.

Given,

$v = 0$

$u = 19.6\ m/s$

$g = -10\ m/s^2$

Substitute these values into equation(1)

$0^{2} - 19.6^{2} =2\times-10\times h_{max}$

$h_{max}= \frac{19.6^{2} }{20} =19.208 \ m$