Question

Two stones are launched from the top of a tall building. One stone is thrown in a direction 15.0 ∘∘ above the horizontal with a speed of 20.0 m/sm/s ; the other is thrown in a direction 15.0 ∘∘ below the horizontal with the same speed.

Which stone spends more time in the air? (Neglet air resistance)


a. The stone thrown upward spends more time in the air.

b. The stone thrown downward spends more time in the air.

c. Both stones spend the same amount of time in the air.

Answer 1

a. The stone thrown upward spends more time in the air.

Step-by-step explanation:

Given:

projection of first stone,
`\theta_1=15^(\circ)` above the horizontal

initial velocity of projectiles,
`u_1=u_2=20\ m.s^(-1)`

projection of second stone,
`\theta_2=15^(\circ)` below the horizontal

The stone thrown upward will spend more time in the air because it travels more distance than the one thrown downwards.

The stone thrown upwards faces deceleration due to the gravity because it goes opposite to the gravity initially, then reaches a velocity zero for a moment and then falls freely from a greater height.

While the second stone posses an initial velocity downward in the direction of the gravity and which further increases its velocity and it travels a short distance.

Answer 2

Answer:a

Step-by-step explanation:

Given

First stone is thrown
`15^(\circ)` above the horizontal with some speed let say u

Second stone is thrown at
`15^(\circ)` below the horizontal with speed u

For a height h of building

For first stone (motion in vertical direction)

using

`v^2-u^2=2ah`

where v=final velocity

u=initial velocity

a=acceleration

h=displacement

`h=u\sin 15(t_1)-(1)/(2)gt_1^2---1`

For second stone

`h=(-u\sin 15)(t_2)-(1)/(2)gt_2^2----2`

Equating 1 and 2

`u\sin 15(t_1+t_2)-(1)/(2)g(t_1-t_2)(t_1+t_2)=0`

`(t_1+t_2)[u\sin 15-4.9(t_1-t_2)]=0`

as
`t_1+t_2` cannot be zero

so
`t_1-t_2=1.05\ s`

`t_1=t_2+1.056`

therefore time taken by first stone(thrown upward) will be more.