The next two questions refer to the situation below. Two balls are thrown with l

Question

The next two questions refer to the situation below.

Two balls are thrown with launch angles of 30° at t = 0. The first ball is thrown down with speed v from the top of a vertical cliff of height ho, while the second ball is thrown up from ground level at the base of the cliff with sped 2v.

Compare the horizontal distances travelled by the balls at the time t = t* defined as the time the second ball reaches its maximum height.

x1(t*) < x2(t*)

x1(t*) = x2(t*)

x1(t*) > x2(t*)

Compare the speeds of the balls at the time t = t* defined as the time the second ball reaches its maximum height.

v1(t*) < v2(t*)

v1(t*) = v2(t*)

v1(t*) > v2(t*)

Explanation / Answer

For first ball : horizontal velocity = V cos30, vertical velocity = - v sin30

for second ball : horizontal velocity = 2V cos30, vertical velocity = 2 v sin30

for second ball, at maximum height final speed = 0
=> 0 = 2 v sin30 - g t
=> t = 2 v sin30/ g = v/g

horizontal distance travlled by ball 1 at t = v/g ::: s = ut + 1/2 at^2
there is no acceleration in horizontal direction.
=> x1 = v cos 30 t
=> x1 = v cos 30 * v/g = v^2 cos30/g

for second ball : x2 = 2v cos 30 v/g = 2v^2 cos30/g

=> x2(t*) > x1(t*) as it is twice of x1

Horizontal speed of ball 1 = v cos 30 = v cos 30 = v sqrt3/2
vertical speed of ball 1 = - v sin 30 - gt = - v sin 30 - g * v/g = - v sin30 - v = -1/2 v - v = -3/2 v

=> speed = sqrt ( 3/4 v^2 + 9/4 v^2) = sqrt ( 3v^2) = v sqrt3

Horizontal speed of ball 2 = 2v cos 30 = 2v sqrt3 / 2 = v sqrt3
vertical speed of ball 2 = 2v sin 30 - gt = 2v*1/2 - g*v/g = v-v = 0

this means both speeds will be same at time t* when second ball reaches it max height

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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