Question

A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20 degrees above the ground from a height of 4.5 ft. Approximately how long does it take for the ball to hit the ground? Acceleration due to gravity is 32 ft./s^2.
A. 2.35 seconds
B. 2.47 seconds
C. 6.46 seconds
D. 6.50 seconds

Answer 1

Answer:

T= 2.35 seconds

Step-by-step explanation:

⇒The question is on the time of flight.

⇒Time of flight is the time taken for a projected object to reach the ground.It depends on the projectile angle and the initial velocity of the projectile

Given;

Initial velocity of ball= 110ft./sec.

The projectile angle= 20°

Acceleration due to gravity, g=32 ft./s²

⇒Formulae for time of fright T= (2×u×sin Ф)/g

Where T=time of fright, u=initial velocity of projectile, Ф=projectile angle and g=acceleration due o gravity.

Substituting values

T= (2×u×sin Ф)/g

T=( 2×110×sin 20°) / 32

T= 2.35 seconds

Answer 2

Answer:

Option B - Time taken for the ball to hit the ground is 2.47 seconds.

Step-by-step explanation:

Given :  A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20 degrees above the ground from a height of 4.5 ft.

To find : How long does it take for the ball to hit the ground?

Solution :

According to question,

The equation that models the height of the ball in feet as a function of time is

$h(t) = h_0 + v_0t -16t^2$

Where, $h_0$ is the initial height,

$v_0$ is the initial velocity and

t is the time in seconds.

We have given,

Initial height, $h_0=4.5 ft.$

A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20 degrees.

The initial speed, $v_0=110\times \sin(206\circ)$

$v_0=37.62ft/s$

We have to find the time for the ball to hit the ground i.e. h(t)=0

Substitute all the values in the formula,

$0 =4.5+ 37.62t -16t^2$

Applying quadratic formula to solve the equation,

The solution of quadratic general equation $ax^2+bc+c=0$ is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where, a=-16 , b=37.62 , c=4.5

Substituting in the formula,

$t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}$

$t=\frac{-37.62\±\sqrt{1703.2644}}{-32}$

$t=\frac{-37.62\±41.270}{-32}$

$t=\frac{-37.62+41.270}{-32},\frac{-37.62-41.270}{-32}$

$t=-0.114,2.47$

neglecting the negative value

t=2.47 seconds

Therefore,Option B is correct.

Time taken for the ball to hit the ground is 2.47 seconds.