Question

NEED HELP ASAP PLS (with work)

A punt is kicked at an angle of 60 with a speed of 40 m/s, and lands at the same
height from which it was kicked.

How far away will the ball land?

What is the maximum height reached by the ball?

What is the hang time? Total

Answer 1

1) The ball will land approximately 141.4 m away from where it was kicked

2) The maximum height reached is approximately 61.22 m

3) The total hang time is approximately 7.07 seconds

Step-by-step explanation:

First we list out the known parameters

The direction in which the punt is kicked, θ = 60°

The speed with which the punk is kicked, v₀ = 40 m/s

1) The distance away (the range, R) the ball will land is given by the equation;

`R = (v_0^2 * sin(2\cdot \theta))/(g)`

Where;

g = The acceleration due to gravity = 9.8 m/s²

Substituting the values, gives;

`R = (40^2 * sin(2*60^(\circ )))/(9.8) \approx 141.4 \ m`

The distance away the ball will land ≈ 141.4 m

2) The maximum height is given m=by the following equation for vertical motion;

(v₀ × sin(θ))² = 2 × g ×
`h_(max)`

Obtained from the equation of motion, v² = u² - 2×g×h where at maximum height;

v = 0 at maximum height, the ball stops upward motion

h = Height of motion =
`h_(max)` at maximum height

v, and u are the vertical component of the velocity, v₀

Substituting the values in the equation for the maximum height, gives;

(40 × sin(60))² = 2 × 9.8 ×
`h_(max)`

(40 × (√3)/2)² = 1200 = 19.6 ×
`h_(max)`

1200 = 19.6 ×
`h_(max)`

`h_(max)` = 1200/19.6 ≈ 61.22 m

The maximum height reached ≈ 61.22 m

3) The time to maximum height is given by the following equation of motion;

`v_y` = v₀ × sin(θ) - g × t

`v_y` = 0 at maximum height

We therefore have;

v₀ × sin(θ) = g × t

t = v₀ × sin(θ)/g = 40 × sin(60°)/9.8 = (40 ×(√3)/2)/9.8 = 3.535 seconds

Given that the motion of the ball is symmetrical about the maximum height, we have;

The time from ground level to maximum height = The time from the maximum height back to the ground level

The total time hang time = The time from ground level to maximum height + The time from the maximum height back to the ground level

The total time hang time ≈ 3.535 + 3.535 ≈ 7.07 seconds.

The total time hang time ≈ 7.07 seconds.