Question

A ball is shot from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = (7.6 + 6.7 ) m/s, with horizontal and upward.
(a) To what maximum height does the ball rise?

(b) What total horizontal distance does the ball travel?

(c) What is the magnitude of the ball's velocity just before it hits the gound?

(d)What is the direction of the ball's velocity just before it hits the ground?

Answer 1

Answer:

(a) 11.34m

(b) 22.89m

(c) 16.87m/s

(d) 63.2°

Explanation:

Given;

v = (7.6 i + 6.7 j) m/s   [where i is horizontal and j is upward]

Since this is a projectile motion, the horizontal and vertical components of the motion are considered;

Using one of the equations of motion;

v² = u² + 2as            ------------------------(i)

Where;

v = final velocity

u = initial velocity

a = acceleration

s = displacement

Resolving equation (i) into the horizontal component

vₓ² = uₓ² + 2aₓsₓ        --------------------(ii)

Where;

vₓ = final velocity of the horizontal motion

uₓ = initial velocity of the horizontal motion

aₓ = acceleration of the horizontal motion

sₓ = horizontal displacement

In the horizontal motion, velocity is constant. i.e

vₓ = uₓ

Therefore, acceleration, aₓ = 0

Substitute these values into equation (ii), we have

=> vₓ² = uₓ² + 2(0)sₓ

=> vₓ² = uₓ²

=> vₓ = uₓ

From the given equation;

v = (7.6 i + 6.7 j) m/s;

Where;

i = horizontal component of the velocity

j = vertical component of the velocity

=> vₓ = uₓ = 7.6 m/s

Resolving equation (i) into the vertical component

$v^{2}_{y}$= $u^{2} _{y}$ + $2a_{y}s_{y}$      --------------------(iii)

Where;

$v_{y}$ = final velocity of the vertical motion

$u_{y}$ = initial velocity of the vertical motion

$a_{y}$ = acceleration of the vertical motion = g

$s_{y}$ = vertical displacement or height

In the vertical motion, acceleration is constant and is equal to the acceleration due to gravity, g = 10m/s². i.e

$a_{y}$ = ±g

Substitute these values into equation (iii), we have

=> $v^{2}_{y}$ = $u^{2} _{y}$ + $2a_{y}s_{y}$

=> $v^{2}_{y}$ = $u^{2} _{y}$ + 2g$s_{y}$       ----------------------(iv)

From the question,

at a height 9.1m, v = (7.6 i + 6.7 j) m/s;

Where;

i = horizontal component of the velocity

j = vertical component of the velocity

=> $s_{y}$ = 9.1, $v_{y}$ = 6.7m/s and g = -10m/s² (since the ball moves upwards against gravity)

Substitute these values into equation (iv) to calculate $u_{y}$ as follows;

6.7² = $u^{2} _{y}$ + 2(-10) x 9.1

44.89 = $u^{2} _{y}$  - 182

$u^{2} _{y}$  = 44.89 +  182

$u^{2} _{y}$  = 226.89

Solve for $u_{y}$;

$u_{y}$ = $\sqrt{226.89}$

$u_{y}$ = 15.06m/s

(a) Using equation (iv);

$v^{2}_{y}$ = $u^{2} _{y}$ + 2g$s_{y}$

At maximum height, $v_{y}$ = 0

g = -10m/s²

Substitute these values and $u_{y}$ = 15.06m/s into the equation as follows;

0 = (15.06)² + 2(-10)$s_{y}$

0 = 226.80  - 20 $s_{y}$

226.80  = 20 $s_{y}$

$s_{y}$ = 226.80 / 20

$s_{y}$ = 11.34 m

Therefore, the maximum height that the ball rises is 11.34m

(b) The horizontal distance sₓ that the ball travels is given by the range formula of a projectile as follows;

sₓ = 2 x $u_{y}$ x uₓ / g

sₓ = 2 x 15.06 x 7.6 / 10

sₓ = 2 x 15.06 x 7.6 / 10

sₓ = 22.89 m

Therefore, the total horizontal distance that the ball travels is 22.89m

(c) The magnitude of the ball's velocity before it hits the ground is the vector sum of the initial velocities of its horizontal and vertical motion and is given by;

|v| = $\sqrt{(u_{x} ^{2} ) + (u_{y} ^{2} )}$

|v| =  $\sqrt{7.6^{2} + 15.06^{2} }$

|v| = $\sqrt{57.76 + 226.80}$

|v| = $\sqrt{284.56}$

|v| = 16.87 m/s

Therefore, the magnitude of the ball's velocity before it hits ground is 16.87m/s

(d) The direction (θ) of the ball's velocity just before it hits the ground is;

θ = tan⁻¹ ($u_{y}$ / uₓ)

θ = tan⁻¹ (15.06 / 7.6)

θ = tan⁻¹ (1.98)

θ = 63.2°

Therefore, the direction of the ball's velocity before it hits ground is 63.2°