physics question

Question

A football is kicked at 16.5m/s at 57 degrees above the ground. How long will the ball stay in the air? What is the maximum height of the ball? Will the ball clear a bar 3.0m high and 20.0m away from the kicker?

Answer 1

Calculate the x- and y-components of the initial speed, vi

v_ix = v_icosθ = (16.5m/s)cos(57°) = 8.99m/s

v_iy = v_isinθ = (16.5m/s)sin(57°) = 13.34m/s

a) total time in air, t_air = ?

use the kinematic equation, v_fy = v_iy +at , but a = -g = -9.8m/s²

==> t = v_iy/g = (13.34m/s)/(9.8m/s²) = 1.36s

The total time of the ball in air is twice this number, i.e t_air = 2.72s

b) using v_fy² = v_iy² + 2aΔy, but at maximum height, v_fy =0

==> Δy = v_iy²/2g = 9.08m [remember a = -g]

c) solve for the horizontal distance travel, using Δx =v_ixt_air = 24.44m compare this to 20.0m location of the bar

and also Δy = 9.08m compared to the 3.0m high bar and make your deduction!

Answer 2

First, find the horizontal and vertical components of the Initial velocity:

Vh_i = 16.5 Cos(57)

Vv_i = 16.5 Sin(57)

First, find the time t₁ to the highest point (where Vv = 0) This is the initial Vv divided by the acceleration of gravity.

The maximum height is then t₁ times the average vertical velocity (8.25)

Then, use the free-fall equation to find the time to fall back to the ground:

d =1/2 * a * t² after the find the "fall" time, add t₁ to get the total time the ball is in the air

For the second question:

first, get the time to travel 20 m, using the initial horizontal velocity. call this t(20)

From the first question, draw a sketch of the height of the ball vs. time. On this sketch mark the given height (3 m) and the time t(20)