A ball starts 25 meters above the ground and is throw meters/second n upward at

Question

A ball starts 25 meters above the ground and is throw meters/second n upward at an initial speed of 32 meters/second (a) How high is the ball above the ground 3.8 seconds after being thrown? (b) Calculate the maximum height of the ball above the ground. A ball is thrown downward with an initial speed of 7.0 meters/second from the roof of a building which is 55 meters tall. (a) Find the time it takes the ball to reach the ground. (b) Find the speed of the ball when it is 20 meters below the roof of the building.

Explanation / Answer

C2 a)initial speed of ball = 32 m/s(vertically upwards)

acceleration on ball = acceleration due to gravity = 9.8m/s2 (vertically downwards)

S = ut+(1/2)at2

h = (32*3.8)+(1/2)*(-9.8)(3.82)

= 50.844m'

Since the ball started 25 meters above the ground, its height after 3,8 seconds measured from ground = 25+50.844 = 75.844meter

b)let h be the maximum height reached by ball

when the ball reached maximum height, it's velocity becomes zero

v2=u2+2as

0 = 322 +(2*-9.8*h)

h = 52.24m

Maximum height reached when measure from ground = 25+52.24 =77.24meters

C3

a)we know, s=ut+(1/2)at2

55 = 7t+(1/2)*9.8*t2

Therefore, t = 2.71 seconds

b)v2=u2+2as

= (72)+(2*9.8*20)=441

v= sqrt(441)=21 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.