Ansteeslook 150 Woses onlyA ball is thrown vertically up words and sis

1.	Ansteeslook 150 Woses onlyA ball is thrown vertically up words and sisto a heightof 126.5 m. calculate thei velocity by which the ball was thrown upwards and2the time taken by the ball to reach the highest point.(Given is g = 985 )
Answer» Answer: Given the height at which ball is THROWN, h = 126.5 m. 1) To calculate the initial velocity of ball by which ball was thrown, use equation of motion, V ^2 - u ^2 = 2 gh Here, u is the initial velocity v is final velocity and h is height at which ball reached. At height 126.5 m (maximum height) , the final velocity is zero. Substitute the given values, we get 0= u ^2 - 2× 9.8m/sec ^2 × 126.5m u = √2× 9.8 m/ sec ^2 × 126.5 u = 49.79m / sec Thus, the velocity by which the ball was thrown is 49.79 m/s. 2) To calculate the time taken by the ball to reach the HIGHEST point, use equation of motion v = u - gt Substitute the values, we get 0 = 49.79 - 9.8 m/ sec ^2 ×t t = 49 .79 m/ sec ÷ 9.8 m/ sec t = 5 .08 sec

Ansteeslook 150 Woses onlyA ball is thrown vertically up words and sisto a heightof 126.5 m. calculate thei velocity by which the ball was thrown upwards and2the time taken by the ball to reach the highest point.(Given is g = 985 )

Answer»

Answer:

Given the height at which ball is THROWN, h = 126.5 m.

1) To calculate the initial velocity of ball by which ball was thrown, use equation of motion,

V ^2 - u ^2 = 2 gh

Here, u is the initial velocity v is final velocity and h is height at which ball reached.

At height 126.5 m (maximum height) , the final velocity is zero.

Substitute the given values, we get

0= u ^2 - 2× 9.8m/sec ^2 × 126.5m

u = √2× 9.8 m/ sec ^2 × 126.5

u = 49.79m / sec

Thus, the velocity by which the ball was thrown is 49.79 m/s.

2) To calculate the time taken by the ball to reach the HIGHEST point, use equation of motion

v = u - gt

Substitute the values, we get

0 = 49.79 - 9.8 m/ sec ^2 ×t

t = 49 .79 m/ sec ÷ 9.8 m/ sec

t = 5 .08 sec