Question

1) A ball is kickedfrom the top of a bridge at a height of 50 m (measured from the surface of the water below) with an initial speed of 30 [m/s] at 22 degrees with respect to the x-axis.

a) Determine the time it takes for the ball to reach its maximum height.

b) Determine the maximum height of the ball (measured from the water below)

c) Determine the speed of the ball at its heighest point on its trajectory

d) Determine the time at which the ball lands/.

e) Where does the ball land? [i.e. Calculate its range]

f) Calculate the velocity of the ball just before it lands[both magnitude and direction]

Answer 1

For vertcal,
u_y = 30 sin(22⁰) m/s
a_y = - 9.8m/s²

For horizontal
u_x = 30 cos(22⁰)
a_x = 0

a) At maximum height, vertical velocity = 0.

So, 0 = 30 sin(22⁰) - 9.8 t
=> t = 1.14675 sec

b) If s is the height travel from the point of release,
0² = (30 sin(22⁰))² - 2 x 9.8 x s
=> s = 6.44m
Maximum height, h = 50 + 6.44 = 56.44 m

c) At highest point,  vertical velocity = 0 and horizontal velocity does not change,

So, speed = u_x = 30 cos(22⁰) = 27.82 m/s

d) Let t be the time at which the ball lands,
Height of the ball is given by,

h = 50 + u_y t + 1/2 at²

h = 0 when the ball lands.

So,
0 = 50 + 30 sin(22⁰) t - 0.5 x 9.8 t²
4.9 t² - 11.24 t - 50 = 0

Solving this gives, t = 4.54 s, -2.25 s
But, out t is +ve here, so

t = 4.54 sec

e) Range = u_x x 4.54 = 27.82 x 4.54 = 126.30m

f) At landing, v_y = u_y + at
=> v_y = 30 sin(22⁰) + (-9.8) x 4.54 = -33.25m/s

v_x = u_x = 27.82 m/s

So, magnitude of velocity,

Angle it makes with the x- axis,