Question

1. A soccer ball is kicked at an upward angle θ between 0° and 90°. Its initial vertical (y) velocity vy0 is 19.6 m/s. It follows a parabolic trajectory and lands at the end of its range.

c. If the ball’s range is 60 m, is the angle θ less than, equal to, or greater than 45°?

2. A small toy car is placed upon a ramp that is 2 m long. The car is released (with zero initial velocity) from the top of the ramp. Assume that friction is negligible.(aramp= 1 m/s2)

How high is the top if the car takes 2.0 s to reach the bottom?

Answer 1

#1

As the ball is projected at some angle with vertical speed

$v_{y0} = 19.6 m/s$

now we can find the time of flight

$T = \frac{2v_{y0}}{g}$

$T = \frac{2*19.6}{9.8}$

$T = 4 s$

now we can find the horizontal speed as

Range = horizontal speed * time

$60 = v_{x0}*T$

$60 = v_{x0}* 4$

$v_{x0} = 15 m/s$

now for the projection angle we will have

$tan\theta = \frac{v_{y0}}{v_{x0}}$

$tan\theta = \frac{19.6}{15}$

so here angle is more than 45 degree as the above ratio is more than 1

#2

here we know that car takes t = 2 s to reach the bottom

here acceleration = 1 m/s^2

initial speed = 0

we can use kinematics now

$d = v_i*t + \frac{1}{2} at^2$

$d = 0 + \frac{1}{2}*1*2^2$

$d = 2 m$

so the toy car is released from the top at distance 2 m from the bottom