Question

A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a hillside 4.00 s later. Neglect air friction. (a) What is the projectile’s velocity at the highest point of its trajectory? (b) What is the straight-line distance from where the projectile was launched to where it hits its target?

Answer 1

Answer:

(a) 51.96m/s

(b) 207.84m

Explanation:

Using one of the equations of motion as follows;

h = ut + $\frac{1}{2}$ at²;

Where;

h = vertical/horizontal displacement of the body in motion;

u = initial velocity of the body in motion

t = time taken for the body to attain the displacement

a = acceleration of the body,

... the motion of the projectile can be resolved into horizontal and vertical components as follows;

(i) Vertical component

$h_{y}$ = $u_{y}$t + $\frac{1}{2}$ $a_{y}$t²;  ----------------(ii)

Where;

$h_{y}$ = vertical displacement of the body in motion;

$u_{y}$ = initial vertical velocity of the body in motion = usinθ [θ is the angle of projection of the body above the horizontal]

t = time taken for the body to attain the vertical displacement

a = acceleration of the body = g [acceleration on the vertical motion is due to gravity]

Substitute the value of $u_{y}$ and $a_{y}$ into equation (ii) as follows;

$h_{y}$ = usinθt + $\frac{1}{2}$ gt²;   ----------------------(iii)

(ii) Horizontal component

$h_{x}$ = $u_{x}$t + $\frac{1}{2}$ $a_{x}$t²;        -----------------(iv)

Where;

$h_{x}$ = horizontal displacement of the body in motion;

$u_{x}$ = initial horizontal velocity of the body in motion = ucosθ [θ is the angle of projection of the body above the horizontal]

t = time taken for the body to attain the horizontal displacement

a = acceleration of the body = 0 [acceleration on the vertical motion is zero due to constant velocity]

Substitute the value of $u_{y}$ and $a_{y}$ into equation (iv) as follows;

$h_{x}$ = ucosθt + $\frac{1}{2}$ (0)t²;

$h_{x}$ = ucosθt;           ----------------------------(v)

(a) As a projectile goes up in the air, the velocity of its vertical component decreases. When the projectile attains its highest point of its trajectory, the vertical component of its velocity becomes zero. However, the horizontal component of the velocity of the projectile remains constant throughout the flight. Therefore, at the projectile's highest point in trajectory, its velocity (v) is given by only the horizontal component as follows;

v = $u_{x}$ = ucosθ                ----------------(vi)

Where;

u = initial velocity of the projectile = 60.0m/s

θ is the angle of projection of the body above the horizontal = 30.0°

Substitute these values into equation (vi) as follows;

$u_{x}$ = 60 x cos30.0°

$u_{x}$ = 60 x 0.8660

$u_{x}$ = 51.96 m/s

Therefore, the projectile’s velocity at the highest point of its trajectory is 51.96m/s

(b) The straight line distance (horizontal distance = $h_{x}$) from where the projectile was launched to where it hits its target is given by equation (v)

$h_{x}$ = ucosθt  

=> $h_{x}$ = $u_{x}$t      [since $u_{x}$ =  u cos θ]                     --------------------(vii)

Where;

t = time taken for the motion = 4.00s

$u_{x}$ = 51.96m/s  [as calculated above]

Substitute these values into equation (vii) as follows;

$h_{x}$ = 51.96 x 4.00

$h_{x}$ = 207.84m

Therefore, the straight-line distance from where the projectile was launched to where it hits its target is 207.84m

Answer 2

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be  

v_ox=v_o*cosФ

       =60*cos (30)

      = 51.96 m/s^-1