An athlete executing a long jump leaves the ground at a 20° angle and travels 7.80 m.

(a) What was the takeoff speed?
1._____ m/s
(b) If this speed were increased by just 5.0%, how much longer would the jump be?
2.______ m

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To find the answers to these questions, we can use the principle of projectile motion. The key here is to break down the initial velocity into its horizontal and vertical components.

(a) To find the takeoff speed, we need to determine the initial velocity of the athlete. We know the angle of takeoff (20°) and the distance traveled (7.80 m).

First, let's find the initial vertical velocity (Vy) using the formula Vy = V * sin(theta), where V is the initial velocity and theta is the angle of takeoff.

Vy = V * sin(20°)

Since the flight time of the jump is the same as the time of descent, we can use the formula:

h = (1/2) * g * t^2

where h is the maximum height of the jump, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight. In this case, t = 2 * t_up.

Using the equation for time of flight:

t = 2 * t_up = 2 * (V * sin(theta) / g)

Therefore, the total time in the air (t_total) can be expressed as:

t_total = t_up + t_down = (V * sin(theta) / g) + (V * sin(theta) / g) = 2 * (V * sin(theta) / g)

Since the horizontal distance traveled is given by:

d = Vx * t_total = V * cos(theta) * t_total

Substituting the values, we get:

d = V * cos(theta) * 2 * (V * sin(theta) / g) = 2 * V^2 * sin(theta) * cos(theta) / g

From this equation, we can solve for V:

V = sqrt((d * g) / (2 * sin(theta) * cos(theta)))

Plugging in the known values:

V = sqrt((7.80 m * 9.8 m/s^2) / (2 * sin(20°) * cos(20°)))

Calculating this, we get:

V ≈ 6.77 m/s

Therefore, the takeoff speed is approximately 6.77 m/s.

(b) To find the increase in distance if the speed is increased by 5.0%, we can use the formula:

new distance = old distance + (5.0% of old distance)

Plugging in the known values:

new distance = 7.80 m + (0.05 * 7.80 m)

Calculating this, we get:

new distance ≈ 8.19 m

Therefore, the jump would be approximately 8.19 m long, which is an increase of 0.39 m.