A ball is thrown upward from the top of a building at an angle of 30.0° to the h

Question

A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 21.0 m/s. The point of release is h = 46.0 m above the ground.

a)How long does it take for the ball to hit the ground?

b)Find the ball's speed at impact

c)Find the horizontal range of the ball

Suppose the ball is thrown from the same height as in the problem abover at an angle of 34.0° below the horizontal. If it strikes the ground 42.2 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)

a)the time of flight

b) the inital speed

c) the speed and angle of the velocity vector with respect to the horizontal at impact

Explanation / Answer

a) Taking UP as the positive direction
s = ut + 0.5 at^2
where:
s = displacement (difference between the initial and final positions) = – 46 metres
u = vertical component of initial velocity = 21 * sin30 = 10.5 m/s
a = acceleration = – 9.8 m/s/s
t = time taken in seconds

Rearranging to give:
0.5at^2 + ut – s = 0
then the quadratic formula gives:
t = – u / a ± (u^2 / a^2 + 2s / a)
= – 10.5 / – 9.8 ± (10.5^2 / (– 9.8)^2 + 2 * – 46 / – 9.8)
= 1.07143±3.24588
= 4.31731 seconds

b) Now, considering velocity downwards i.e. DOWN as the positive direction
h = (u * sin )^2 / 2a
where:
h = peak height of ball trajectory in metres
u = initial velocity = 21 m/s
= firing angle = 30 degrees
a = acceleration = 9.8 m/s/s

h = (21 * sin30)^2 / (2 * 9.8)
= 5.625 m

v^2 = u^2 + 2as
where:
v = final velocity in m/s
u = initial velocity = 0 m/s as ball is stationary momentarily at peak height
s = displacement from peak height to ground = 5.625 + 46 = 51.625 m

v = (2 * a * s)
= (2 * 9.8 * 51.625)
=31.80 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.