A person throws a ball with a velocity of 37.0 m/s at an angle of 25.0degrees ab

Question

A person throws a ball with a velocity of 37.0 m/s at an angle of 25.0degrees above the horizontal. The ball is released at a height of 1.00 m above a level horizontal field. For each of the following, write the information you have, the variable you need, and the equation(s) you are using - show all work... What is the time of flight of the ball? What is the horizontal distance traveled by the ball when it first hits the ground? What is the speed of the ball just before it hits the ground? What is the maximum height the ball achieves (above the ground) during its flight?

Explanation / Answer

Initial Horizontal Velocity = 37 * cos(25) = 33.5 m/s
Initial Vertical Velocity = 37 * sin(25) = 15.6 m/s

Horizontal Acceleration = 0
Vertical Acceleration = 9.8 m/s^2 downwards

Time of Flight = Time taken by ball to reach max height + Time taken by ball to reach ground.
Time taken by ball to reach max height -
At Max height, V = 0

V = u - gt
t1 = 15.6/9.8
t1 = 1.59 s

Max Height Reached =
V^2 = u^2 - 2gh
h = u^2/2g
h = (15.6)^2/(2*9.8)
h = 12.4 m

Total Height = 12.4 + 1 = 13.4 m

Time taken by ball to reach ground =
S = u*t + 0.5 at^2
13.4 = 0 + 0.5*9.8*t^2
t2 = sqrt(13.4/4.9)
t2 = 1.65 s

Time of Flight = t1 + t2
Time of Flight = 1.59 + 1.65
Time of Flight = 3.24 s

b)
Horizontal distance travelled by ball = 33.5 * 3.24
Horizontal distance travelled by ball = 108.5 m

c)
Speed of Ball =
Vertical Speed V = u + a*t
V = 0 + 9.8 * 1.65
V =  16.17 m/s

Horizontal speed = 33.5

Speed of Ball = sqrt(33.5^2 + 16.17^2)
Speed of Ball just before it hits the ground = 37.2 m/s

d)
Maximum height ball achieves = 13.4 m

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