To determine
To find: The velocity after 2 s and after 4 s if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.
Answer
The velocity after 2 s is 4.9 m/s.
The velocity after 4 s is −14.7 m/s.
Explanation
The height of the particle is, h(t)=2+24.5t−4.9t2.
Find the derivative of h(t) and obtain the velocity.
v(t)=h'(t)=24.5−9.8t
Thus, the velocity is v(t)=24.5−9.8t (1)
Substitute t = 2 in equation (1) and obtain the velocity after 2 s.
v(2)=24.5−9.8(2)=24.5−19.6=4.9
Thus, the velocity after 2 s is 4.9 m/s.
Substitute t = 4 in equation (1) and obtain the velocity after 4 s.
v(4)=24.5−9.8(4)=24.5−39.2=−14.7
Thus, the velocity after 2 s is −14.7 m/s.
To determine
To find: when the projectile reach its maximum if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.
Answer
The projectile reaches its maximum height at 2.5 s.
Explanation
From part (a) the velocity of the particle is v(t)=24.5−9.8t.
Set v(t)=0 and obtain the t value.
24.5−9.8t=09.8t=24.5t=2.5
Thus, the projectile reaches its maximum height at t = 2.5 s.
To determine
To find: when the projectile reach its maximum if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.
Answer
The maximum height of the projectile is 32.625 m.
Explanation
From part (b), the projectile reaches its maximum height at t = 2.5 s.
Substitute t = 2.5 s for h(t) and obtain the maximum height of the projectile.
h(2.5)=2+24.5(2.5)−4.9(2.5)2=2+61.25−30.625=32.625
Thus, the maximum height of the projectile is 32.625 m.
To determine
To find: When the projectile hits the ground if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.
Answer
The projectile hits the ground approximately at t = 5.08 s.
Explanation
The projectile hits the ground at h(t)=0.
Set h(t)=0 and solve for t.
2+24.5t−4.9t2=04.9t2−24.5t−2=0
Use quadratic formula and solve further as follows.
t=24.5±(−24.5)2−4(4.9)(−2)2(4.9)=24.5±600.25+39.22(4.9)=−24.5±25.289.8≈5.08 or −0.07
Ignore the negative value, the value of t is approximately 5.08.
Thus, the projectile hits the ball when t= 5.08 s.
To determine
To find: With what velocity the projectile hits the ground if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.
Answer
The projectile hits the ground with the velocity −25.3 m/s.
Explanation
From part (d), the projectile hits the ground at h(5.08).
Substitute t = 5.08 for v(t) and obtain the required velocity.
v(5.08)=24.5−9.8(5.08)=24.5−49.784=−25.284
Thus, the projectile hits the ground with the velocity of −25.284 m/s.