An elevated catapult launches a ball with a velocity of 28.0 m/s, directed at an

Question

An elevated catapult launches a ball with a velocity of 28.0 m/s, directed at an angle of 48° above the horizontal. The catapult is elevated vertically 6.0 m above the ground.

(a) What is the horizontal component of the ball's velocity when it launches?
m/s

(b) What is the vertical component of the ball's launch velocity?
m/s

(c) When does the ball land?
s

(d) How far away does the ball land, as measured horizontally from the ground under the catapult?
m

(e) At what time does the ball reach its highest point?
s

(f) When the ball is at its highest point, how far is it, as measured horizontally from the catapult?
m

(g) When the ball is at its highest point, how high is it relative to the ground?
m

Explanation / Answer

a) vix = vicos= 28cos48 = 18.74m/s

b) viy = visin =28sin48 = 20.8m/s

c) h= viy*t – 1/2gt^2

-6.0 = 20.8*t - 1/2*9.8*t^2    => t= 4.52s

d) X = vix*t = 28.74*4.52 = 129.9 m

e) At Hmax , vfy= 0

vfy = viy –gt

0 = 20.8 – 9.8*t     => t= 2.12s

f) X= vix* = 18.74*2.12 = 39.73m

g) H = viy*t -1/2gt^2 = 20.8*2.12 – ½*9.8*2.12^2 = 22.07m

Hmax = 6.0 + 22.07 = 28.07m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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