Let's analyze the given projectile motion equation:

$h(t) = -4.9t^2 + v_0 t + h_0$

where:

• $v_0 = 24.5$ m/s (initial velocity),
• $h_0 = 7$ m (initial height),
• Acceleration due to gravity is $-9.8$ m/s².

Step 1: Find Time to Reach Maximum Height

At maximum height, the velocity becomes zero. The velocity function is obtained by differentiating $h(t)$:

$v(t) = \frac{dh}{dt} = -9.8t + v_0$

Setting $v(t) = 0$:

$0 = -9.8t + 24.5$

Solving for $t$:

$t = \frac{24.5}{9.8} = 2.5 \text{ seconds}$

Step 2: Find Maximum Height

Substituting $t = 2.5$ into the height equation:

$h(2.5) = -4.9(2.5)^2 + 24.5(2.5) + 7$

$= -4.9(6.25) + 24.5(2.5) + 7$

$= -30.625 + 61.25 + 7$

$= 37.625 \text{ meters}$

Final Answer:

• Time to reach max height: 2.5 seconds
• Maximum height: 37.625 meters

Would you like a breakdown of each calculation step?

Related Questions:

1. What is the velocity of the object when it hits the ground?
2. How long does it take for the object to hit the ground?
3. What is the total flight time of the projectile?
4. How would the maximum height change if the initial velocity were doubled?
5. How does air resistance affect the motion of the projectile?

Tip: The maximum height in projectile motion depends only on the initial vertical velocity and gravity, not the object's mass.