A particle is projected vertically upwards from its ground with an initial speed of [tex]$30 ms^{-1}$[/tex]. Calculate the:
1. Maximum height reached by the particle.
2. Time taken by the particle to return to the ground.
3. Time taken for the particle to attain a height of 4 m above its ground. Take g = [tex]$10 ms^{-2}$[/tex]
Answer (1)
Calculate the maximum height using v 2 = u 2 + 2 a s , where v = 0 , u = 30 m/s , and a = − 10 m/s 2 , resulting in a maximum height of 45 m .
Determine the total time to return to the ground using s = u t + 2 1 a t 2 , where s = 0 , u = 30 m/s , and a = − 10 m/s 2 , giving a total time of 6 s .
Find the times to reach 40 m using s = u t + 2 1 a t 2 , where s = 40 m , u = 30 m/s , and a = − 10 m/s 2 , which yields times of 2 s and 4 s .
The maximum height is 45 m , the total time is 6 s , and the times to reach 40 m are 2 s and 4 s .
Explanation
Problem Setup We are given that a particle is projected vertically upwards with an initial speed of 30 m/s . We need to find the maximum height reached, the total time to return to the ground, and the time(s) to reach a height of 40 m . We are also given that g = 10 m/s 2 .
Maximum Height Calculation To find the maximum height, we use the kinematic equation v 2 = u 2 + 2 a s , where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. At the maximum height, the final velocity v = 0 . The acceleration is due to gravity, so a = − g = − 10 m/s 2 . Thus, we have
0 = ( 30 m/s ) 2 + 2 ( − 10 m/s 2 ) s
0 = 900 − 20 s
20 s = 900
s = 20 900 = 45 m .
Therefore, the maximum height reached by the particle is 45 m .
Total Time Calculation To find the total time taken for the particle to return to the ground, we use the kinematic equation s = u t + 2 1 a t 2 , where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Since the particle returns to the ground, the displacement s = 0 . Thus, we have
0 = 30 t + 2 1 ( − 10 ) t 2
0 = 30 t − 5 t 2
0 = t ( 30 − 5 t )
So, t = 0 or 30 − 5 t = 0 . The solution t = 0 corresponds to the initial time when the particle is launched. The other solution is
5 t = 30
t = 5 30 = 6 s .
Therefore, the total time taken for the particle to return to the ground is 6 s .
Time to Reach 40m To find the time(s) taken for the particle to attain a height of 40 m , we use the same kinematic equation s = u t + 2 1 a t 2 , where s = 40 m . Thus, we have
40 = 30 t + 2 1 ( − 10 ) t 2
40 = 30 t − 5 t 2
5 t 2 − 30 t + 40 = 0
Dividing by 5, we get
t 2 − 6 t + 8 = 0
Factoring the quadratic equation, we have
( t − 2 ) ( t − 4 ) = 0
So, t = 2 s or t = 4 s .
Therefore, the particle attains a height of 40 m at t = 2 s (on the way up) and at t = 4 s (on the way down).
Final Answer The maximum height reached by the particle is 45 m , the time taken for the particle to return to the ground is 6 s , and the times taken for the particle to attain a height of 40 m are 2 s and 4 s .
Examples
Understanding projectile motion is crucial in many real-world scenarios. For example, engineers use these principles to design safe and effective amusement park rides, ensuring that riders experience the thrill of motion while remaining secure. Similarly, in sports, athletes and coaches use projectile motion concepts to optimize performance in activities like basketball, where understanding the trajectory of the ball can improve shooting accuracy.
