Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

16. peregrine falcons dive on their prey with speeds up to 36 m/s. from…

Question

  1. peregrine falcons dive on their prey with speeds up to 36 m/s. from what height must a falcon fall freely to achieve this speed?
  2. minnie fay the ballerina leaps straight upward with a speed of 1.8 m/s. how long is she in the air? (again, let’s call “up” the positive direction so the acceleration due to gravity is negative (down). also, what is the final speed at the top?)
  3. in an incident on the m1 motorway in england, a jaguar sports car made a skid 276 m long while braking to a stop. assuming a constant deceleration of -8 m/s², what was the car’s initial speed?

Explanation:

Response
Problem 16

Step1: Identify knowns and formula

We know the initial velocity \( v_i = 0 \, \text{m/s} \) (starts from rest), final velocity \( v_f = 36 \, \text{m/s} \), and acceleration \( a = g = 9.8 \, \text{m/s}^2 \) (free - fall acceleration). The kinematic equation that relates \( v_i \), \( v_f \), \( a \), and displacement \( d \) (which is the height \( h \) we want to find) is \( v_f^2=v_i^2 + 2ad \).

Step2: Rearrange the formula to solve for \( d \)

Since \( v_i = 0 \), the equation simplifies to \( v_f^2=2ad \), and we can solve for \( d \) (height \( h \)): \( h=\frac{v_f^2}{2g} \)

Step3: Substitute the values

Substitute \( v_f = 36 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \) into the formula:
\( h=\frac{36^{2}}{2\times9.8}=\frac{1296}{19.6}\approx66.12 \, \text{m} \)

Step1: Analyze the motion

The ballerina leaps upward with initial velocity \( v_i = 1.8 \, \text{m/s} \), acceleration \( a=-g=- 9.8 \, \text{m/s}^2 \) (negative because it's in the opposite direction of the positive upward direction), and when she lands, her displacement \( y - y_0=0 \) (starts and ends at the same vertical position). The kinematic equation for displacement is \( y - y_0=v_it+\frac{1}{2}at^{2} \)

Step2: Substitute values into the equation

Substitute \( y - y_0 = 0 \), \( v_i=1.8 \, \text{m/s} \), and \( a=- 9.8 \, \text{m/s}^2 \) into \( y - y_0=v_it+\frac{1}{2}at^{2} \):
\( 0 = 1.8t+\frac{1}{2}\times(-9.8)t^{2} \)
Factor out \( t \): \( 0=t(1.8 - 4.9t) \)
We have two solutions for \( t \): \( t = 0 \) (initial time) and \( 1.8-4.9t=0 \)

Step3: Solve for \( t \) from \( 1.8 - 4.9t=0 \)

\( 4.9t=1.8 \)
\( t=\frac{1.8}{4.9}\approx0.37 \, \text{s} \) (time to go up and come back down, so total time in air is approximately \( 0.37 \, \text{s} \))

Part 2: Final speed at the top

Step1: Analyze the motion at the top

At the top of the motion, the vertical velocity \( v_f = 0 \, \text{m/s} \) (because the ballerina momentarily stops before starting to fall back down). This is a characteristic of projectile motion (or vertical motion under gravity) at the maximum height - the vertical component of velocity is zero.

Answer:

The falcon must fall from a height of approximately \( \boldsymbol{66.1 \, \text{m}} \) (or more precisely \( \boldsymbol{\frac{1296}{19.6}\approx66.12 \, \text{m}} \))

Problem 17
Part 1: Time in the air