An object is launched at a velocity of 36m/s in a direction making an angle of 29⁰ upward with the horizontal
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the magnitude of the velocity of the object just before it hits the ground?
Answers
Given : An object is launched at a velocity of 36m/s in a direction making an angle of 29⁰ upward with the horizontal
To Find :
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the magnitude of the velocity of the object just before it hits the ground?
Solution:
Horizontal Velocity = 36cos29° = 31.49 m/s
Vertical Velocity = 36Sin29° = 17.45 m/s
Horizontal Velocity remains constant
Vertical velocity becomes 0 at max height
V = u + at
a = - g = -9.8 m/s²
=> 0 = 17.45 - 9.8t
=> t = 1.78 sec
Total Time of flight = 2 * 1.78 = 3.56 sec
V² - U² = 2as
Max height = (v² - u²)/2a = -(17.45)²/(2*(-9.8))
= 15.53 m
magnitude of the velocity of the object just before it hits the ground
= 36 m/s ( based on conservation of energy )
Initial KE + initial PE = Final KE + Final PE
Initial and final PE = 0
Initial KE = Final KE
Hence Magnitude of velocity will be same
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