A ball is launched from ground level, as in the simulation above, with initial v

Question

A ball is launched from ground level, as in the simulation above, with initial velocity components of 41.0 m/s horizontally and 34.0 m/s vertically. Use g = 10 m/s2, and neglect air resistance.

(a) Exactly 2.00 seconds after launch, what is the angle of the balls velocity, measured from the horizontal?
(b) Exactly 2.00 seconds after launch, how high above the ground is the ball?
(c) What is the maximum height reached by the ball?
(d) What is the balls time-of-flight? In other words, how much time passes between the launch and the ball reaching the ground again?  

Please explain how to do this :(

Explanation / Answer

a. Horizontal velocity remains same as there is no acceleration horizontally

vertical velocity after 2 seconds = intitial vertical velocity + acceleration due to gravity * time

= 34 - 10 * 2 = 34 - 20 = 14

tan (theta) = vertical velocity / horizontal velocity = 14 / 41

=> theta = taninverse(14 / 41) = 18.85 degrees

b. vertical distance travelled in 2 seconds = 34 * 2 - 0.5 * 10 * 2^2

[since, from formula s = ut + 1/2 at^2]

= 68 - 20 = 48 meters

c. maximum height reached by the ball h = v^2 / 2 * g = 34 * 34 / 2 * 10 = 57.8 m

d. time of flight = 2 * time taken by the ball to reach maximum height

time taken by the ball to reach maximum height = v / g = 34 / 10

time of flight = 34 * 2 / 10 = 6.8 seconds

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