Projectile motion

Question

A ball is thrown from near the ground at an angle of 45 degrees with the ground. The initial speed is 40 m/s.

a) What is the magnitude and direction of acceleration when the ball reaches it's max height?

b) What is the highest that the ball goes?

c) What is the horizontal range?

Answer 1

A projectile is in free fall so the only force on it is the force due to gravity downwards (normally we ignore air resistance)

So from F=ma, there is no acceleration horizontally, and acceleration 9.8 m/s/s downwards (constant)

So at all points in the flight acceleration is 9.8m/s/s downwards (answering a)

b) °The initial vertical speed of the ball is 40 sin (45°°)

When at the maximum height, speed is instantaneously zero.

v² = u² + 2as

v = 0, u = 40 sin (45°), a = -9.8. using this you can calculate s, the maximum height

c) Range = v² sin (2 x 45) / g, where v= 40 m/s, g = 9.8

Answer 2

x = 40cos45°t; v_x = 40cos45°;

y = 40sin45°t - 4.9t²; v_y = 40sin45°- 9.8t;

a) accelaretion a = - 9.8 m/s²(gravity)

b)At highest point v_y = 40sin45° - 9.8t = 0; t = 20√2/9.8 = 2.89 s

max h = y(2.89) = 20√2·2.89 - 4.9·2.89² = 40.82 m

c) Horizontal range: at the landing y = 20√2t - 4.9t² = 0; t = 20√2/4.9 = 5.77 s

s = 20√2·5.77 = 163.2 m