Question

a ball is thrown vertically upwards with a velocity of 49m/s
1 the maximum height to which it ries
2the total time it takes to return to the surface of the earth​

Answer 1

Answer:

source;google

Explanation:

Given parameters

Initial velocity of the ball (u) = 49m/s.

The velocity of the ball at maximum height (v) = 0.

g = 9.8m/s2

Let us considered the time taken is t to reach the maximum height H.

Consider a formula,

2gH = v2 – u2

2 × (- 9.8) × H = 0 – (49)2

– 19.6 H = – 2401

H = 122.5 m

Now consider a formula,

v = u + g × t

0 = 49 + (- 9.8) × t

– 49 = – 9.8t

t = 5 sec

(1) The maximum height to ball rises = 122.5 m

(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.

hope this helps

With love,

Eshaal    ;]

Answer 2

Concept :- When a body is thrown vertically upward then its final Velocity is Zero(0). It has initial Velocity only. But when a body is projected at some angle then it has final Velocity which becomes zero at the highest point and after that ball tends to return.

In this question,

Let the height attained by body = h

v = 0 m/s

u = 49 m/s

g = 10 m/s² (Considering upward acceleration as negative)

Since, time is not given. Hence, we will apply 3rd Law of Motion.

v² - u² = 2gh

-(49)² = 2 × (-10) h

-(49 × 49)/-20 = h

h = 122.5 m

Again,

Total time taken by body to attain height of 122.5 m = T

v = u + at

0 = 49 - 10t ( Acceleration is negative)

-49 = -10t

t = 4.9 s

Total time = 2t ( Ball is returning also).

T = 2 × 4.9 s

T = 9.8 s