Question

A ball is thrown vertically upwards and reaches a maximum height in 3 seconds. Find:
(1) the velocity with which it was thrown upwards
(2) the maximum height attained by the ball.

(I WILL MARK BRAINLIEST FOR THE ON WHO ANSWERS 1ST!!)

Answer 1

(1)= ,v=u+at
0=u-9.8×3
0=u-29.4
u=29.4m/s
(2)
v²=u²+2as
0=(29.4)²+2×(-9.8)×s
0=864.36-19.6s
-864.36=-19.6s
-864.36÷(-19.6)=s
s=44.1m

Answer 2

Given,

Time (t) = 3 seconds

Final Velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 $m/s^{2}$

To Find,

1) The velocity with which it was thrown upwards

2)The maximum height attained by the ball

Solution,

1) We know the kinematic equation,

     $v = u + at$          

Using this equation, we will get the value of the initial velocity

     $0 = u + (9.8*3)$

     $u =-( 9.8*4)$

     $u = -29.4 m/s^{2}$

The negative sign shows direction of the initial velocity is opposite the gravity acceleration.

Therefore, The velocity with which it was thrown upwards is 29.4 m/s2

2) To find the maximum height,

We will use the formula,

     $v^{2}=u^{2} + 2as$      [Here, s is the height]

Substituting the values we get,

     $0 = (-29.4)^2 + 2* 9.8*s$

     $0 = 864.36 + 19.6s$

Solving this equation we get,

     s = - 44.1 m

As the distance can not be negative we will consider only the magnitude.

Therefore,

The maximum height attained by the ball is 44.1 m

1) The velocity with which it was thrown upwards is 29.4 m/s2

2) The maximum height attained by the ball is 44.1 m