Question

A stone is catapulted at time t = 0, with an initial velocity of magnitude 17.0 m/s and at an angle of 48.0° above the horizontal. (Neglect air resistance.) Find its horizontal and vertical displacements from the catapult site at the following times after launch.(a) 1.10 s? m (horizontal)? m (vertical)(b) 1.70 s? m (horizontal)? m (vertical)(c) 2.59 s? m (horizontal)? m (vertical)

Answer 1

a)x= 12.50 m,y= 7.84 m

b)x= 19.32 m,y= 7.02 m

c)x= 29.44 m,y= -0.81 m

Step-by-step explanation:

Given that

Initial velocity U= 17 m/s

θ = 48°

This is the case of projectile motion

So the horizontal component of velocity U = U cosθ

u = U cosθ

u = 17 cos 48°

u=11.37 m/s

The vertical component of velocity U = U sinθ

Vo= U sinθ

Vo= 17 sin 48°

Vo = 12.63 m/s

a)

t= 1.1 s

We know that in projectile motion horizontal component of velocity will remain constant.

So the horizontal distance ,x

x = u .t

x = 11.37 x 1.1 m

x= 12.50 m

In vertical direction

`y=V_ot-(1)/(2)gt^2`

`y=12.63* 1.1-(1)/(2)* 10* (1.1)^2`

y= 7.84 m

b)

t= 1.7 s

So the horizontal distance ,x

x = u .t

x = 11.37 x 1.7 m

x= 19.32 m

In vertical direction

`y=V_ot-(1)/(2)gt^2`

`y=12.63* 1.7-(1)/(2)* 10* (1.7)^2`

y= 7.02 m

c)

t= 2.59 s

So the horizontal distance ,x

x = u .t

x = 11.37 x 2.59 m

x= 29.44 m

In vertical direction

`y=V_ot-(1)/(2)gt^2`

`y=12.63* 2.59-(1)/(2)* 10* (2.59)^2`

y= -0.81 m