A person is throwing two balls in the air one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls every second, how high do they rise?

Question

A person is throwing two balls in the air one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls every second, how high do they rise? (A) 5 m (B) 3.75 m (C) 2.50 m (D) 1.25 m

Tardigrade Admin · Accepted Answer

Time of ascent =1 s ⇒ (u/g)=1 ⇒ u=10 ms -1 Maximum height attained =(u2/2 g)=(102/2 × 10) =5 m

Q. A person is throwing two balls in the air one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls every second, how high do they rise?

5 m

3.75 m

2.50 m

1.25 m

Time of ascent $= 1 s$
$\Rightarrow \frac{u}{g} = 1$
$\Rightarrow u = 10 m s^{- 1}$
Maximum height attained $= \frac{u ^{2}}{2 g} = \frac{1 0 ^{2}}{2 \times 10}$
$= 5 m$

Q. A person is throwing two balls in the air one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls every second, how high do they rise?

5 m

3.75 m

2.50 m

1.25 m

Time of ascent =1s ⇒gu​=1 ⇒u=10ms−1 Maximum height attained =2gu2​=2×10102​ =5m

Time of ascent $= 1 s$
$\Rightarrow \frac{u}{g} = 1$
$\Rightarrow u = 10 m s^{- 1}$
Maximum height attained $= \frac{u ^{2}}{2 g} = \frac{1 0 ^{2}}{2 \times 10}$
$= 5 m$