Question

Can anyone please work out the projectile motion questions?

Answer 1

a) `v_(0x)=17.809ms^-1`
b) `y_max=6.318m`
c) `T=2.271s`
d) `R=40.446m`
e) `v_y=-0.6317ms^-1`

Explanation:

Given:
Initial velocity of projection: `v_0=21ms^-1`
Angle of projection `theta=32^@`

a)
Horizontal component of the initial velocity

`v_(0x)=v_0costheta`

`v_0costheta=21cos32^@`

`21cos32^@=17.809ms^-1`

Thus,

`v_(0x)=17.809ms^-1`

b) Maximum height reached by the ball

Vertical component of the initial velocity

`v_(0y)=v_0sintheta`

`v_0sintheta=21sin32^@`

`21sin32^@=11.128ms^-1`

`v_(0y)=11.128ms^-1`

`y_max=(v_(0y))^2/(2g)`

`g=9.8ms^-2`

`y_max=11.128^2/(2xx9.8)`

`y_max=6.318m`

c) Time of flight of the soccer ball.

initial vertical velocity is

`v_(0y)=11.128ms^-1`

final vertical velocity is

`-v_(0y)=-11.128ms^-1`

Change in velocity is

Final velocity - initial velocity

`=-11.128-11.128`

`delv_y=-22.256`

Acceleration due to gravity is
`g=9.8ms^-2`

Acceleration during the journey is

`a_y=-9.8`

Time of flight is

`T=(delv_y)/a_y=-22.256/-9.8`

`T=2.271s`

d) Range of the soccer ball

Acceleration along horizontal direction is zero

`R=v_(0x)T`

`v_(0x)=17.809ms^-1`

`T=2.271s`

`R=17.809xx2.271`

Range of the soccer ball is

`R=40.446m`

e) Velocity of the soccer ball after time t = 1.2 s

`v_x=v_(0x)`
Same for the entire journey

`v_y=v_(0y)+a_yt`

`v_y=11.128+(-9.8)t`

`v_y=11.128-9.8t`

Substitute t=1.2s

`v_y=11.128-9.8xx1.2`

`v_y=-0.6317ms^-1`