A ball with mass m kg is thrown upward with initial velocity 19 m/s from the roof of a building 24 m high. Neglect air resistance. Use g = 9.8 m/s². Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. Xmax= meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. 1 end = i S

QUESTION 2 A ball is thrown straight up in the air. When it reaches a height that is n times smaller than the maximum height it can reach, its speed is v. Calculate the maximum height, in meters. n = 3.9 v = 16.8 m/s to saue all answers.

3. An object is dropped from a height h above the surface of the Earth such that it experiences an initial acceleration of 0.76g. Find the height h above the surface of the Earth. ME a = G (R, +h)

I need part d and e

What is her velocity at 15 s?

A bb is shot downward from the top of a building leaving the barrel with an initial speed of 90.3 m/s downward. The top of the building is 60.1 m above the ground. How much time elapses between the instant it leaves the barrel and the instant of impact with the ground? Round your answer to 3 decimal places.

3. Bob (52.0kg) was curious about the amount of air resistance (friction) that would act on him as he fell through the air, so he conducted an experiment. He stepped out of a third story window, 15.0m above the ground, and measured that it took 1.94s to hit the ground (don't worry, there was a soft mat to cushion the impact). Assuming his initial velocity was zero as he stepped out of the window, determine the force of air friction that acted on Bob as he fell.

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Finding x(t) and v(t) by Integration 5. The acceleration of a jeepney is given by a, (t) = at – ßt? т where a = 1.50 m/.3 and ß = 0.120 m /s4 The motorcycle is at rest at the origin when time t = 0. a. Find its position and velocity as functions of time. b. Calculate the maximum velocity it attains.

olem 1. A stone is dropped from the top of a building and falls freely from rest. What is its position (height) and its velocity after 1.0s, 2.0s, and 3.0s? To find for the position (height), we will use: Know that initial velocity (vi) = 0 a. Find h after 1.0s h=0(1.0s)-(9.8 m/s)(1.0s)2 2 |h=v₁t - 9² 2 h=0-9.8m 2 h=-4.9 m (negative sign denotes downward direction) b. Find h after 2.0s h=0(2.0s) - (9.8 m/s²)(2.0s)² 2 h=0-39.2 m 2 h = -19.6 m or 19.6 m downward Given: v₁ = 20 m/s Find: a. v after 1s b. v after 5s c. d after 1s d. t from origin to maximum height d from origin to maximum height e. Solutions: a. y after is use: V₁ = v₁ - gt vy=20 m/s-(9.8 m/s²)(1.0s) vy= 20 m/s -9.8 m/s vy=10.2 m/s b. v after 2s use: Sample Problem 2. A boy threw a ball upward with an initial velocity of 20 m/s and was able to catch it before it reached the ground on its return. a. What was the ball's velocity after 1s? b. What was the ball's velocity after 2s? c. What was its displacement after 1s? d. How…