Question

A ball is thrown vertically upward . It has a speed of 10ms^-1 when it has reached one half of its maximum height . How high does the ball rise ? Take g = 10 ms^-1

Answer 1

Answer:

The ball rises 10 metre high.

Explanation:

When a ball is thrown upwards with a speed u, the maximum height it can gain= $\bold{\frac{u^{2}}{2 g}}$

g is the gravitational force with a value of $\bold{10 m s^{-2}}$

At one half of the maximum height, the speed of the ball is 10m/s

We use the equation:

$v^{2}=u^{2}-2 g H$ (This is because the ball is thrown upwards with some velocity and a retardation is produced due to gravitation of the earth and so we take the value of g=-g).

0=102-2×10×H/2  

0=100- 10H  

10H=100  

H=10m

Answer 2

Answer: 10meter

Explanation:

Maximum height = h=

2g

u

2

​

.......1

, u being the initial velocity

using third equation v

2

−u

2

=2g×height gained =2g×

2

h

​

=gh

putting u

2

=2gh in above equation from first equation we get v

2

−2gh=gh or h=

3g

v

2

​

=

10

10×10

​

=10meter