Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

An object is launched at a velocity of 20 / in a direction
making an angle of 25° upward with the horizontal. What is the
maximum height reached by the object? What is the total flight time
(between launch and touching the ground) of the object? What is the
horizontal range (maximum x above ground) of the object? What is the
magnitude of the velocity of the object just before it hits the

Sagot :

bharathparasad577

When an object is thrown with an angle with horizontal, it makes projectile motion

It is given that angle of projectile was 25 degree

The formulas of projectile motion is as follows:

component of velocity at time t = Vxcos[tex]\alpha[/tex]

time of maximum  height= vsin[tex]\alpha[/tex] /g

maximum height = (v sin[tex]\alpha[/tex] )^2/2g

horizontal range = v^2 sin2[tex]\alpha[/tex]/g

v is given to be 20 m/s

time of flight is given by v 2sin[tex]\alpha[/tex]/g

therefore, time of flight = 20 x 2x0.4226 /10 = 1.690 seconds

maximum horizontal range = 20 x 20 x 0.766/10 =30.641 meters

Disclaimer: The value of acceleration due to gravity is assumed to be 10 m/s^2 also the speed of velocity has no unit and hence m/s has been considered

#SPJ9

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.