When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of 11.3 m/s, at an angle of 28° above the horizontal.

(a) What is the range of his leap?
(b) For how much time is he in the air?

To find the range of the greyhound's leap, we need to determine the horizontal distance it travels. We can use the horizontal component of its velocity to calculate this.

(a) To find the horizontal component of velocity, we use the formula:

Vx = V * cos(θ)

where:
Vx is the horizontal component of the velocity,
V is the initial velocity (11.3 m/s), and
θ is the angle above the horizontal (28°).

Vx = 11.3 m/s * cos(28°)
Vx ≈ 10.047 m/s

Now, we can find the range by calculating the time the greyhound is in the air and multiplying it by the horizontal component of velocity.

(b) To find the time the greyhound is in the air, we can use the formula:

t = 2 * Vy / g

where:
Vy is the vertical component of the velocity,
g is the acceleration due to gravity (approximately 9.8 m/s²).

To find Vy, we use the formula:

Vy = V * sin(θ)

Vy = 11.3 m/s * sin(28°)
Vy ≈ 5.111 m/s

Now, we can calculate the time:

t = 2 * 5.111 m/s / 9.8 m/s²
t ≈ 1.05 s

Finally, we can find the range by multiplying the time by the horizontal component of velocity:

Range = Vx * t
Range ≈ 10.047 m/s * 1.05 s
Range ≈ 10.549 m

Therefore, the range of the greyhound's leap is approximately 10.549 meters, and it is in the air for approximately 1.05 seconds.

4.09

Use the initial vertical velocity to determine the time in air. From that, determine the range. I will be happy to critique your thinking or work