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

Question

A. 2.35 seconds
B. 2.47 seconds
C. 6.46 seconds
D. 6.50 seconds

Answer 1

The time it takes for the ball to hit the ground can be calculated using the kinematic equation:

s = ut + (1/2)at^2

where:
s = displacement (height) = -4.5 ft (negative because the ball is falling)
u = initial velocity = 110 ft./sec
a = acceleration due to gravity = 32 ft./s^2
t = time

Using the given angle of 20°, we can find the vertical component of the initial velocity:

V_vertical = 110 ft./sec * sin(20°) = 110 ft./sec * 0.342 = 37.62 ft./sec

Now, we can use the vertical component of the velocity to find the time it takes for the ball to hit the ground:

-4.5 ft = 37.62 ft./sec * t + (1/2)(-32 ft./s^2)t^2
-4.5 = 37.62t - 16t^2
16t^2 - 37.62t - 4.5 = 0

Using the quadratic formula, we get:

t = (37.62 ± sqrt((37.62)^2 - 4*16*(-4.5))) / 32
t = (37.62 ± sqrt(1415.8244 + 288)) / 32
t = (37.62 ± sqrt(1703.8244)) / 32
t = (37.62 ± 41.29) / 32

Since we are looking for the positive time, we choose the solution with the plus sign:

t = (37.62 + 41.29) / 32
t = 2.47 seconds

Therefore, it takes approximately 2.47 seconds for the ball to hit the ground.

B. 2.47 seconds